Two students are on a balcony 18.3 m above the street. One student throws a ball (ball 1) vertically downward at 12.2 m/s; at the same instant, the other student throws a ball (ball 2) vertically upward at the same speed. The second ball just misses the balcony on the way down.
(a) What is the difference in the two ball's time in the air?
(b) What is the velocity of each ball as it strikes the ground?
ball 1
ball 2
(c) How far apart are the balls 0.800 s after they are thrown?
cannot seem to figure out how to find part a) i cannot get the second balls time.
a. d1 = 12.2t + 0.5gt^2 = 18.3m.
12.2t + 0.5*9.8t^2 = 18.3,
4.9t^2 + 12.2t - 18.3 = 0,
Solve using Quadratic Formula and get:
t1 = 1.054, and -3.54s.
Use positive value of t:
T1 = 1.054s = Time in air.
Vf = Vo + gt,
t2(up) = (Vf-Vo) / g,
t2(up) = (0-12.2) / -9.8 = 1.24s.
Vf^2 = Vo^2 + 2gd,
d2(up) = (Vf^2-Vo^2) / 2g,
d2(up) = (0-148.8) / -19.6 = 7.59m.
D2 = 7.59 + 18.3 = 25.89m from ground.
d = Vo*t + 0.5gt^2 = 25.89,
0 + 0.5*9.8*t^2 = 25.89,
4.9t^2 = 25.89,
t^2 = 5.28,
t(down) = 2.3s.
T2 = 1.24 + 2.3 = 3.54s in air.
T2 - T1 = 3.54 - 1.054s difference.
b. Vf^2 = Vo^2 + 2gd,
V1^2 = 12.2 + 2*9.8*18.3 = 370.88,
V1 = 19.26m/s.
V2^2 = 0 + 2*9.8*25.89 = 507.44,
V2 = 22.53m/s.
c. d1 = Vo*t + 0.5gt^2
d1 = 12.2*0.8 + 0.5*9.8*(0.8)^2 = 12.90m below the balcony.
d2=12.2*0.8 + 0.5*(-9.8)(0.8)^2 = 6.62m above the balcony
D = 6.62 + 12.90 = 19.52m apart.
A particle (a) is projected horizontally at 50m/s from the top of a tower 100m high at same instant aoother particle is projected from bottom of the tower in same vertical plain at 100m/s with elevation 60^0.after how maoy seconds will the particle collide
To find the difference in the two balls' time in the air (part a), we need to calculate the time it takes for each ball to hit the ground.
For the first ball (ball 1), since it is thrown directly downward, we can use the equation for vertical motion:
h = vit + (1/2)gt^2
Where:
- h is the initial height (18.3 m)
- vi is the initial velocity (12.2 m/s)
- g is the acceleration due to gravity (-9.8 m/s^2, taking into account the direction)
Rearranging the equation to solve for time (t):
t = √((2h) / g)
Substituting the given values:
t1 = √((2 * 18.3) / 9.8) = √(36.6 / 9.8) = √3.7347 ≈ 1.932 s
Therefore, the first ball (ball 1) will take approximately 1.932 seconds to hit the ground.
Now, for the second ball (ball 2), we can apply the same equation. However, the initial velocity is positive (12.2 m/s) because it is thrown vertically upward:
t2 = √((2 * h) / g)
t2 = √((2 * 18.3) / 9.8) ≈ 1.932 s
Both balls have the same time in the air, which is approximately 1.932 seconds.
Moving on to part b, we need to find the velocity of each ball as it strikes the ground.
For ball 1, since it is thrown downward, its final velocity is the same as its initial velocity (12.2 m/s), but negative due to the downward direction.
Therefore, the velocity of ball 1 (V1) as it strikes the ground: V1 = -12.2 m/s.
For ball 2, since it is thrown upward, we can use the equation for final velocity in vertical motion:
vf = vi + gt
Where:
- vi is the initial velocity (12.2 m/s, upward)
- g is the acceleration due to gravity (-9.8 m/s^2, downward)
Therefore, the velocity of ball 2 (V2) as it strikes the ground can be calculated as follows:
V2 = 12.2 - 9.8 * t2
V2 = 12.2 - 9.8 * 1.932
V2 ≈ 12.2 - 18.996
V2 ≈ -6.796 m/s
Thus, the velocity of ball 2 (ball thrown upward) as it strikes the ground is approximately -6.796 m/s.
Moving on to part c, we need to find the distance between the balls 0.800 seconds after they are thrown.
To calculate this, we need to determine the distance traveled by each ball during this time frame.
For ball 1 (thrown downward):
The distance traveled by ball 1 (D1) can be calculated using the equation for vertical motion:
D1 = vi * t1 + (1/2) * g * t1^2
D1 = -12.2 * 0.800 + (1/2) * (-9.8) * (0.800)^2
D1 ≈ -9.76 + (-3.136)
D1 ≈ -12.896 m
For ball 2 (thrown upward):
The distance traveled by ball 2 (D2) can also be calculated using the equation for vertical motion:
D2 = vi * t2 + (1/2) * g * t2^2
D2 = 12.2 * 0.800 + (1/2) * (-9.8) * (0.800)^2
D2 ≈ 9.76 + (-3.136)
D2 ≈ 6.624 m
Therefore, the balls are approximately 12.896 meters apart 0.800 seconds after they are thrown.