Wednesday

July 29, 2015

July 29, 2015

Total # Posts: 10,769

**physics**

D = Xo * t = 800 Ft. Xo * 5 = 800 Xo = 160 Ft/s. Tr + Tf = 5 s. Tr = Tf. Tr+Tr = 5. Tr = 2.5 s. = Rise time. Y = Yo + g*Tr = 0. Yo = -g*Tr = -(-32)*2.5 = 80 Ft/s. 1. Tan A = Yo/Xo = 80/160 = 0.50. A = 26.6o. 2. Vo = Xo + Yoi = 160 + 80i = 179Ft/s[26.6o]. 3. h = Yo*Tr + 0.5g*Tr...
*July 28, 2015*

**Physics**

(140-700) = -560 rev/min -560rev/min * 1min/60s * 6.28rad/rev = -58.6 rad/s. = The change in velocity. 60s/560rev * 50rev = 5.36 s. a = (-58.6)/5.36 = -10.9 rad/s^2.
*July 28, 2015*

**math**

V = 9 * 16 * sqrt(25) = 720 Cubic units.
*July 27, 2015*

**Physics**

a = u*g = 0.55 * (-9.8) = -5.39 m/s^2. V^2 = Vo^2 + 2a*d. Vo^2 = V^2 - 2a*d. = 0 - 2(-5.39)*52 = 561. Vo = 23.7 m/s.
*July 26, 2015*

**maths**

7/56 = 10/N. 7N = 560. N = 80.
*July 26, 2015*

**physics**

Wt. = M*g.
*July 25, 2015*

**mathematics**

P = Po(1+r)^n. Po = $1,024. r = 0.07. n = 1comp./yr. * 3yrs. = 3 Compounding periods. Solve for P. I = P-Po.
*July 25, 2015*

**physics**

db = 10*Log I2/I1 = 10*Log I2/6 = 20. Log I2/6 = 2. I2/6 = 10^2 = 100. I2 = 600 W/cm^2.
*July 24, 2015*

**College algebra**

2001: t = 0. P(t) = 1500*(2^)^0 = 1500*(1) = 1500 Thousands. Year 2008: t = 2008-2001 = 7 yrs.
*July 24, 2015*

**College Algebra**

X, (X+1), (X+2). x + (x+1) + (x+2) = 966. 3x + 3 = 966. 3x = 963. X = 321. X+1 = 322. X+2 = 323.
*July 23, 2015*

**Unisa**

Suppose the arc subteding the central angle theta has length 20 pi metres and the radius is 24 metres, then what is theta measured in radians
*July 23, 2015*

**Physics**

At what angle are the ropes pulling?
*July 22, 2015*

**Math**

15 = 3*5. 18 = 3*3*2. 27 = 3*3*3. GCF = 3x^12.
*July 22, 2015*

**Physics**

M1*V1 + M2*V2 = M1*V + M2*V. 7900*5 + 1650*20[30o+180o] = 7900V+1650V 39500 + 33,000[210o] = 9550V 39,500 -28,579 - 16,500i = 9550V. 10,921 - 16,500i = 9550V. 19,787[-56.5o] = 9550V. V = 2.07m/s.[-56.5o] S. of E. = 2.07m/s[304o] CCW from +x-axis.
*July 22, 2015*

**math**

Scale Change = (48-44)in/4cm = 1in/cm.
*July 22, 2015*

**math**

P = Po*(1+r)^n. Po = $3,000. r = (3%/4)/100% = 0.0075 = Quarterly % rate expressed as a decimal. n = 4Comp./yr. * 5yrs. = 20 Compounding periods. Solve for P. I = P-Po.
*July 21, 2015*

**math**

See previous post: Fri, 7-17-15, 11:46 AM.
*July 20, 2015*

**Physics**

T1*Cos(180-35) + T2*Cos50 = -36*Cos270. -0.819T1 + 0.643T2 = 0. T1 = 0.785T2. T1*sin(180-35) + T2*sin50 = -36*sin270. 0.574T1 + 0.766T2 = 36. Replace T1 with 0.785T2: 0.574*0.785T2 + 0.766T2 = 36. 0.451T2 + 0.766T2 = 36. 1.22T2 = 36. T2 = 29.6 N. T1 = 0.785T2 = 0.785 * 29.6 = ...
*July 19, 2015*

**Calculus**

Can someone show how this question is solved. Consider the curve given by the equation 2y^3 + y^2 - y^5 = x^4 -2x^3 +x^2 Find all points at which the tangent line to the curve is horizontal or vertical. Thanks!
*July 18, 2015*

**Math**

Length = L1. Width = L1/4. L2 = L1+10. W2 = L1/4 + 6. A2 = A1 + 128 Ft^2. A2 = (L1*L1/4) + 128. L2*W2 = (L1*L1/4) + 128. (L1+10)*(L1/4) + 6) = (L1*L1/4)+128. Multiply both sides by 4: (L1+10)*(L1)+24 = (L1*L1) + 512. L1^2+10L1+24 = L1^2 + 512. L1^2 - L1^2 + 10L1 + 24 = 512. ...
*July 17, 2015*

**math**

8/x = 2. X = 4 = Scale factor. Area = 8/4 * 12/4 = 6 Ft^2.
*July 17, 2015*

**math**

Area = 7*12 * 5*12 = 5,040 Ft^2
*July 17, 2015*

**Intermediate algebra**

post it.
*July 15, 2015*

**Math**

P(6, 4), m = 5. Y = mx + b. 4 = 5*6 + b. b = -26. Y = 5x - 26.
*July 15, 2015*

**science,**

4400N./0.02m^2 = 220,000 N./m^2. F = 220,000N/m^2 * 0.04m^2 = 8800 N.
*July 15, 2015*

**physics**

h = Vo*t + 0.5g*t^2. h = 2*2.5 + 4.9*2.5^2 = 35.63 m.
*July 15, 2015*

**Geometry**

L = Length of a side. A = Apothem = Altitude. L/A = L/L*sin60 = 1/sin60 = 1.1547. So the length of a side is approximately 1.15 times the apothem.
*July 13, 2015*

**science**

Unbalanced force.
*July 13, 2015*

**Physics**

500[90o] + 400[90o+60o] + F3 = 0 500i -346.41+200i + F3 = 0. -346.41 + 700i + F3 = 0. F3 = 346.41 - 700i = 781N.[-63.7o] = 781N.[63.7o] S. of E.
*July 13, 2015*

**physics**

See previous post: Sun, 7-12-15, 10:24 PM.
*July 12, 2015*

**physics**

See previous post: Sun, 7-12-15, 10:24 PM.
*July 12, 2015*

**physics**

W=0.5M*V2^2-0.5M*V1^2 = 0.5M(V2^2-V1^2)= 450(9^2-36^2) = -546,750 J. mg*h = -546,750. 8820h = 546,750. h = 62. m. 2. W = 450(36^2-9^2) = -546,750 J. Note: The negative sign means the system is doing the work.
*July 12, 2015*

**Physics**

Ti*sin(180-65) + T2*sin65 = -Wp*sin270. 1.3*0.90631 + 1.3*90,631 = Wp. Wp = 2.36 N. = Wt. of picture.
*July 12, 2015*

**Physics**

1a. 1.3*Cos(180-65) = -T2*Cos65. -0.549 = -0.423T2. T2 = 1.30 N. The Tensions are equal because the angles are equal.
*July 12, 2015*

**Physics**

1. Are the two strings tied to the same point on the wall? 2. a = u*g = 0.45 * (-9.8)=-4.41 m/s^2. V^2 = Vo^2 + 2a*d. V = 0. Vo = 7.75 m/s. a = 4.41 m/s^2. Solve for d.
*July 12, 2015*

**Calculus**

Thank You very much, I understand it now
*July 12, 2015*

**Calculus**

y = g(x) = cos(x) Can someone show how to estimate g'(pi/2) using the limit definition of the derivative and different values of h. Thanks!
*July 12, 2015*

**physics**

Glad I could help.
*July 11, 2015*

**physics**

4a. Mass = 1*10^-6m^3 * 19,300kg/m^3 = 0.0193 Kg . 4b. Fb = 1*10^-6m^3 * 1000Kg/m^3 = 0.001 Kg. = 0.0098 N. 5. V = 0.25 * 0.5 * 1 = 0.125 m^3. a. Fb = 0.125m^3 * 1000Kg/m^3 = 125 kg. = 1225 N. b. M = 0.125m^3 * 8,600kg/m^3 = c. = Wb = M*g Newtons. 6. V*D = 13 N. V*810 = 13. ...
*July 11, 2015*

**physics**

Glad I could help.
*July 11, 2015*

**physics**

P = E*I = 120 * 5 = 600 Watts = 600 J/s. Energy = P*t = 600J/s * 55s = 33,000 J.
*July 11, 2015*

**MATH**

You did not give the y-intercept nor the point. So I am going to solve a problem with that INFO given. Given: Y-intercept = 4 or(0,4), and P2(2,6). Find the Eq. of the line. P1(0,4), P2(2,6). Slope = (y2-y1)/x2-x1) = (6-4)/(2-0) = 1. Y = mx + b. Y = x + 4.
*July 10, 2015*

**Mathematics**

From what I understood in the problem, the required is the total travel time of car X from town A to town B. In the 180minutes, I think that the "80minutes later" is already included so no need to add it up.
*July 9, 2015*

**Mathematics**

Let x = speed of car X y = speed of car Y L = distance of AB "meet 80minutes later" L = 80x + 80y (distance = velocity * time) The time for car X to reach town B is: tx = ty + 36 But, tx = L/x and ty = L/y Substitute, tx = ty + 36 [80(x+y)]/x = [80(x+y)]/y + 36 ...
*July 9, 2015*

**Mathematics**

A car X left town A for town B at the same time that another car Y left town B for town A, each travelling at constant speed. They met 80 minutes later and car X arrived at town B 36 minutes after car Y reached town A. How long did it take car X to reach town B?
*July 9, 2015*

**Physics**

Work = The change in kinetic energy: Work = KE2-KE1 = 0.5*M*V2^2 - 0.5*M*V1^2 = 31*5^2 - 31*2^2 = 651 J. Work = (Fe-30)*d = 651. (Fe-30)*25 = 651. 25Fe - 750 = 651. 25Fe = 1401. Fe = 56 N. = Force exerted.
*July 9, 2015*

**physics**

M*g = 95 * 9.8 = 931 N. = Wt. of load. Fp = 931*sin50 = 713.2 N. = Force parallel to the incline. Fn = 931*Cos50 = 598.4 N. = Normal = Force perpendicular to the incline. Fk = u*Fn = 0.10 * 598.4 = 59.84 N. = Force of kinetic friction. a. Work = Fk*d = 59.84 * 30 = 1795 N. b. ...
*July 9, 2015*

**Physics**

Correction: b. D1 + D2 = 30+300 = 330 m. 0.5*1.17*t^2 + 0.5*1.1*t^2 = 330. 0.585t^2 + 0.55t^2 = 330. 1.135t^2 = 330. t^2 = 290.75. t = 17.1 s.
*July 9, 2015*

**Physics**

D1 = Distance traveled by train on left. D2 = Distance traveled by train on right a. D1 + D2 = 30 m. 0.5a1*t^2 + 0.5a2*t^2 = 30 0.5*1.17*t^2 + 0.5*1.1*t^2 = 30 0.585t^2 + 0.55t^2 = 30. 1.135t^2 = 30. t^2 = 26.43 t = 5.14s. D1 = 0.5*a1*t^2. a1 = 1.17 m/s^2. t = 5.14 s. Solve ...
*July 9, 2015*

**Algebra**

F(x) = Y = -8x^2 +3x +7 = 0 Use Quadratic Formula: X = (-B +/- sqrt(B^2-4AC))/2A X = (-3 +/- sqrt(9+224))/-16 X = (-3 +- sqrt233)/-16. X = (-3 +- 15.26)/-16 = -0.767 and 1.14.
*July 8, 2015*

**Calculus**

Can someone provide a function that satisfies all of the following: • p(−1) = 3 and lim x→−1 p(x) = 2 • p(0) = 1 and p'(0) = 0 • lim x→1 p(x) = p(1) and p'(1) does not exist Thanks!
*July 8, 2015*

**ITT**

PE = 20J. at top of incline. PE = 10J. halfway down the incline. KE + PE = 20 KE + 10 = 20 KE = 10J. KE = 0.5M*V^2 = 10 M = 1kg. Solve for V.
*July 8, 2015*

**Math**

Diagram: Draw a rectangle with the vert. sides representing the ht. of the Metro Bldg.(h1).and hor. sides are the distance between bldgs.(45m). Draw a diagonal from lower left to upper right and notice that the alternate interior angles are equal(56o).Extend the vert. side on ...
*July 7, 2015*

**Physics**

M*g = 60 * 9.8 = 588 N. = Wt. of the person. Eq1: T1*Cos9.4 + T2*Cos35.3 = 0. T1*Cos9.4 = -T2*Cos35.3. T1 = -0.827T2. Eq2: T1*sin9.4 + T2*sin35.3 =-588[270o]. 0.1633T1 + 0.578T2 = 0 + 588 Replace T1 with -0.827T2: 0.1633*(-0.827T2) + 0.578T2 = 588 -0.135T2 + 0.578T2 = 588 0....
*July 7, 2015*

**physics 20**

See Related Questions: Tue, 2-7-12, 11:59 AM.
*July 6, 2015*

**Physics**

Yo = 15*sin30 = 7.5 m/s. = Ver. component of initial velocity. Y^2 = Yo^2 + 2g*h = 0. h = -Yo^2/2g = -(7.5)^2/-19.6 = 2.87 m. above the roof. V^2 = Vo^2 + 2g*(h+40) = 0 +19.6*42.87 = 840.25 V = 29 m/s.
*July 4, 2015*

**physics**

Glad I could help.
*July 3, 2015*

**physics**

1. Work = 0.5M*V^2 = 0.5*1500*20^2 = 300,000 J. 2. hp * 746W/hp = 1492 Watts=1492J/s. P = F*V = Mg * V = 1492 J/s. 200*9.8 * V = 1492 1960V = 1492 V = 0.76 m/s.
*July 3, 2015*

**Math**

There is no difference between the number of males or females who go to their primary care physician for an annual exam.
*July 3, 2015*

**Math**

2. P = Po*r*t/(1-(1+r)^-t) Po = 9500 - 800 = $8700. r = (9%/12)/100% = 0.0075 = Monthly % rate expressed as a decimal. P = (8700*0.0075*48)/(1-1.0075)^-48 = $10,392. P/t = 10,392/48 = $216.50/mo. 3. P = Po + I = Po + Po*r*t. P = 2419 + 2419*0.052*36 = $2796.36. P/t = 2796.36/...
*July 2, 2015*

**physics**

See Related Questions: Tue, 11-23-2010, 5:05 AM.
*June 29, 2015*

**Physics**

d1 = Vo*t1 + 0.5g*t1^2 = 59.7 m. 0 + 4.9t1^2 = 59.7. t1^2 = 12.18. t1 = 3.49 s. V1 = Vo + g*t1 = 0 + 9.8*3.49 = 34.2 m/s. V2^2 = V1^2 + 2a*d2 = 3.41^2 = 11.63. 34.2^2 - 3.2d2 = 11.63. -3.2d2 = 11.63-1169.64 = -1158. d2 = 362 m. d2 = V1*t2 + 0.5a*t2^2 = 362. 34.2*t2 - 0.80*t2^2...
*June 29, 2015*

**physics**

M*g = 45 * 9.8 = 441 N. = Wt. of trunk. Fn = 441 + F*sin33. Fs = u*Fn = 0.55(441+F*sin33) = 242.6 + 0.3F. (F-Fs) = M*a. (F-(242.6+0.3F) = M*0. F - 242.6 - 0.3 = 0 0.7F = 242.6. F = 347 N.
*June 23, 2015*

**physics**

M1*V1 + M2*V2 = M1*V3 + M2*V4 1100*22 + 2300*0 = 1100*(-11) + 2300*V4 24,200 = -12,100 + 2300V4 2300V4 = 36,300 V4 = +15.8 m/s = Velocity of the truck.
*June 19, 2015*

**physics**

Correct answer = 92%
*June 19, 2015*

**physics**

496/1127 = 0.44 = 44% sin A = 0.85/2.1. A = 23.9o. Fp = 1127*sin23.9 = 457 N. = Force parallel to the ramp. Fn = 1127*Cos23.9 = 1,030 N. = Force perpendicular to the ramp = Normal force. Eff. = Fp/Fap = 457/496 = 0.92 = 92%
*June 19, 2015*

**physics**

T1[26.5o] and T2[131.5o], CCW. T1*Cos26.5 + T2*Cos131.5 = 0 T1*Cos26.5 = -T2*Cos131.5 T1 = 0.74T2 T1*sin26.5 + T2*sin131.5 + 115*sin270 = 0. Replace T1 with 0.74T2: 0.74T2*sin26.5 + T2*sin131.5 + 115*sin270 = 0. 0.33T2 + 0.749T2 - 115 = 0 1.079T2 = 115. T2 = 106.6 N. T1 = ....
*June 17, 2015*

**Physics**

a. V^2 = Vo^2 + 2g*h = 0 h = -Vo^2/2g = -225/-19.6 = 11.5 m. b. V = Vo + g*Tr = 0. Tr = -Vo/g = -15/-9.8 = 1.53 s. = Rise time. 0.5g*Tf^2 = 11.5-5 = 6.5 m. 4.9Tf^2 = 6.5 Tf^2 = 1.33 Tf = 1.15 s. = Fall time. Tr+Tf = 1.53 + 1.15 = 2.68 s. To catch the ball.
*June 17, 2015*

**Physics**

d1 = 0.5a*T1^2 = 45 m. Solve for T1. 0.5*0.88*T1^2 = 45. T1^2 = 102.3 T1 = 10.11 s. to run the 1st 45 m. V = a*T1 = 0.88*10.11 = 8.9 m/s. d2 = V*T2 = 8.9 * T2 = 100-45 = 55 m. T2 = 55/8.9 = 6.18 s. To run the last 55 m. T = T1+T2 = 10.11 + 6.18 = 16.29 s. to run the race.
*June 17, 2015*

**Physics**

a = (Fap-Fk)/M = (25-10)/5 3 m/s^2. d = 0.5a*t^2.
*June 17, 2015*

**math help please**

Po = 575-25 = $550 = Amount of loan. I = Po*r*t = 550*0.15*2yrs = $165. P = Po + I = 550 + 165 = $715 = Total cost. P/t = 715/24mo. = $29.79/mo.
*June 17, 2015*

**science**

F = ma. a = F/m = -2500/1125 = -2.22 m/s^2.
*June 17, 2015*

**physics**

V^2 = Vo^2 + 2a*d = 0 10^2 + 2a*0.04 = 0 2a*0.04 = -100 a = -1250 m/s. Note: The negative sign means that the acceleration opposes the motion.
*June 17, 2015*

**secondary**

Do you mean: y+2x/4 = x/3 or (y+2x)/4 = x/3 y + 2x/4 = x/3 Multiply both sides by 12: 12y + 6x = 4x. 2x + 12y = 0. m = -A/B = -2/12 = -1/6. (y+2x)/4 = x/3. Multiply both sides by 12: 3y + 6x = 4x. 2x + 3y = 0 m = -A/B = -2/3.
*June 16, 2015*

**furthe math**

Parallel lines have equal slopes(gradients): m1 = m2 = -3.
*June 16, 2015*

**further mathematic**

Parallel lines have equal gradients(slopes); m1 = m2 = -3.
*June 16, 2015*

**Science**

M*g = 9.5 N. M*9.8 = 9.5 M = 0.969 kg. Wt. = M*g = 0.969 * 1.64 = 1.58 N.
*June 16, 2015*

**Maths**

Area = 364/16 = 22.75 m^2 Width = W meters. Length = W+3 meters. Area = (W+3) * W = 22.75 W^2 + 3W - 22.75 = 0 Use Quadratic Formula: W = 3.5 meters = Width.
*June 15, 2015*

**basic electronics**

R3 = R Ohms. R2 = 2R Ohms. R1 = 2R/3 Ohms. 1/R1 + 1/R2 + 1/R3 = 1/1000 1/(2R/3) + 1/2R + 1/R = 1/1000 3/2R + 1/2R + 1/R = 1/1000 3/2R + 1/2R + 2/2R = 1/1000. 6/2R = 1/1000. 3/R = 1/1000. R/3 = 1000. R = 3000 Ohms. R1 = 2R/3 = 6000/3 = 2000 Ohms. R2 = 2R = 2*3000 = 6000 Ohms. ...
*June 14, 2015*

**Physics**

15t = 5 t = 0.333h = 20 min. 11.6t = 5 t = 0.431h = 26 min. T = 26 - 20 = 6 min. waiting time.
*June 12, 2015*

**Physics**

Y^2 = Yo^2 + 2g*h = 0 Yo^2 - 19.6*0.0674 = 0 Yo^2 = 1.32 Yo = 1.15 m/s = Ver. component of initial velocity. Yo = Vo*sin50 = 1.15 Vo = 1.15/sin50 = 1.50 m/s = Initial velocity. Xo = Vo*Cos50 = 1.50*Cos50 = 0.964 m/s = Hor. component of initial velocity. Dx = Xo*T = 1.06 m. 0....
*June 12, 2015*

**Science**

1. F max = 20 + 20 = 40 N. @ 0o. 2. The velocity is zero and not changing; therefore, the acceleration is zero also. F = M*a = M*0 = 0.
*June 12, 2015*

**Science**

20 + 20 = 40 N. @ 0o. Answer = 0o.
*June 12, 2015*

**physics**

Correction: Vo = 49 m/s.
*June 11, 2015*

**physics**

First Stone: d = 0.5g*t^2 = 35 m. 4.9t^2 = 35 t^2 = 7.14 t = 2.67 s Second Stone: t = 2.67-2.0 = 0.67 s. to travel 35 m. d = Vo*t + 0.5g*t^2 d = Vo*0.67 + 4.9*0.67^2 = 35 Vo*0.67 = 35 - 2.2 = 32.8 Vo = 49 m
*June 11, 2015*

**science**

P = 10*Cos30 = 8.66 N. Q = 10*Cos(90-30) = 5 N.
*June 8, 2015*

**Math**

Cos x = -0.40 X = 113.6o
*June 7, 2015*

**physics**

Fr = ((Vs+Vt)/(Vs-Vd))*Fd Fr = ((343+42.6)/(343-0))*95300Hz = 107,136 Hz = Freq. reflected.
*June 5, 2015*

**Physics**

V^2 = Vo^2 + 2g*h = 1 10^2 - 19.6h = 1 -19.6h = -99 h = 5.05 m.
*June 2, 2015*

**physics**

M*g = 25kg * 9.8N/kg = 245 N. = Wt. of block. = Normal(Fn). Fap = 120 N. = Force applied. Fk = u*Fn = 0.34 * 245 = 83.3 N. a = (Fap-Fk)/M m/s^2.
*June 2, 2015*

**Digital Circuits**

b. Limit current.
*May 31, 2015*

**geometry**

(8,-1), (5,-2), (x,y) 5-8 = 1/3(x-8) -9 = x-8 X = -1 -2-(-1)) = 1/3(y-(-1)) -1 = 1/3(y+1) -3 = y+1 Y = -4
*May 31, 2015*

**geometry**

(8,-1), (5,-2), (x,y) x-5 = 1/3(x-8) 3x-15 = x-8 2x = 7 X = 3.5 y-(-2) = 1/3(y-(-1)) y+2 = 1/3(y+1) 3y+6 = y+1 2y = -5 Y = -2.5
*May 31, 2015*

**mechanics of dynamic**

55km/h * 2h = 110 km Head start. d1 = d2 110 + 55*t = 80t -25t = -110 t = 4.4 Hours.
*May 31, 2015*

**math**

Tan 45 = h/d1 h = d1*Tan45 Tan55 = h/(d1-40) h = (d1-40)*Tan55 h = d1*Tan45 = (d1-40)*Tan55 d1*1 = (d1-40)*Tan55 Divide both sides by Tan55: 0.7d1 = d1-40 0.3d1 = 40 d1 = 133.3 Ft. h = d1*Tan45 = 133.3 * 1 = 133.3 Ft.
*May 30, 2015*

**physics**

Tan A = Y/X = 35/-12 = -2.91667 A = -71.1o = 71.1o N. of W.
*May 30, 2015*

**college geometry**

Another Method: a. X = (3+5)/2 = 4 Y = (4+1)/2 = 2.5
*May 28, 2015*

**college geometry**

a. A(3,4), C(x,y), B(5,1). Points A and B are equidistance from C: x-3 = 5-x X = 4 y-4 = 1-y Y = 2 1/2 b. A(0,2), C(x,y), B(0,6). x-0 = 0-x X = 0 y-2 = 6-y Y = 4 c. Same procedure as a, and b. d. A(2a,0), C(x,y), B(0,2b). x-2a = 0-x 2x = 2a X = a y-0 = 2b-y 2y = 2b Y = b
*May 28, 2015*

**science**

F = M*a a = F/M = 52.2/3.47 = 15 m/s^2 V = Vo - a*t V = 0 Vo = 17.4 m/s a = 15 m/s^2 t = ?
*May 28, 2015*

**Trig/Geometry**

Tan 15 = h/d Tan 15 = 145/d d = 145/Tan 15 = 541.2 Ft. Or if the angle between the line of sight and the ver.(145Ft) is 15o, then Tan 15 = d/145 d = 145*Tan 15 = 38.85 Ft.
*May 28, 2015*