A ball is thrown with a slingshot at a velocity of 95 ft./sec. at an angle of 25 degrees above the ground from the height of 6 ft. Approximately how long does it take for the ball to hit the ground? Acceleration due to gravity is 32 ft./s^2.
A. 2.65 seconds
B. 2.51 seconds
C. 5.45 seconds
D. 5.38 seconds
Vo = 95ft/s[25o]
Yo = 95*sin25 = 40.1 Ft./s.
Y = Yo+gTr = 0
40.1+(-32)Tr = 0
Tr = 1.25 s. = Rise time.
Y^2 = Yo^2+2g*h = 0
40.1^2+(-64)h = 0
h = 25.1 Ft.
h max = 6+25.1 = 31.5
h max = 0.5g*Tf^2 = 31.5
16Tf^2 = 31.5
Tf = 1.40 s. = Fall time.
Tr+Tf = 1.25+1.40 = 2.65 s. To hit gnd.
half of Unknown Dude's answers are wrong, wouldn't recommend using them
Well, well, well, looks like we have a gravity-fueled ball throwing situation here! And lucky for you, Clown Bot is here to provide some hilarious calculations.
Now, to figure out the time it takes for the ball to hit the ground, we'll need to break down the initial velocity into its horizontal and vertical components. The horizontal component will remain constant, while the vertical component will be affected by gravity.
First, let's find the initial vertical velocity (v0y). We can use basic trigonometry for that. We know the initial velocity is 95 ft./sec., and the angle is 25 degrees. So, v0y = 95 * sin(25).
Next, let's find the time it takes for the ball to reach the ground. We can use the following equation: t = (-v0y ± √(v0y² + 2 * g * h)) / g. Where g is the acceleration due to gravity and h is the initial height of the ball.
Using the given values, we get: t = (-v0y ± √(v0y² + 2 * g * h)) / g = (-v0y ± √(v0y² + 2 * 32 * 6)) / 32.
After plugging in the values, we get two possible solutions for time (due to the ± sign):
t = (-v0y + √(v0y² + 2 * g * h)) / g, and
t = (-v0y - √(v0y² + 2 * g * h)) / g.
Now, let's crunch some numbers, shall we? Using my trusty calculator, I've calculated that the approximate time it takes for the ball to hit the ground is 2.65 seconds.
So, my dear friend, the answer you're looking for is A. 2.65 seconds! Time flies when you're having fun with Clown Bot!
To find the time it takes for the ball to hit the ground, we can use the equations of motion in projectile motion.
First, we need to split the initial velocity into its horizontal and vertical components. The vertical component of velocity can be found by multiplying the initial velocity (95 ft./sec.) by the sine of the launch angle (25 degrees):
Vertical velocity (V_y) = 95 ft./sec. × sin(25)
The horizontal component of velocity can be found by multiplying the initial velocity (95 ft./sec.) by the cosine of the launch angle (25 degrees):
Horizontal velocity (V_x) = 95 ft./sec. × cos(25)
Next, we can use the equation for vertical displacement to find the time it takes for the ball to reach the ground. The initial vertical displacement is given as 6 ft., and the final vertical displacement is 0 ft.:
Final vertical displacement (Δy) = 0 ft. - 6 ft.
The equation for vertical displacement is:
Δy = V_y × t - (1/2) × g × t^2
where t is the time in seconds and g is the acceleration due to gravity (32 ft./s^2).
Plugging in the known values, we get:
0 - 6 = (95 ft./sec. × sin(25)) × t - (1/2) × 32 ft./s^2 × t^2
Simplifying the equation, we have:
-6 = 50t - 16t^2
Rearranging the equation, we get a quadratic equation:
16t^2 - 50t - 6 = 0
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
where a = 16, b = -50, and c = -6, we can solve for t.
Calculating the values within the square root, we have:
√(b^2 - 4ac) = √((-50)^2 - 4(16)(-6)) = √(2500 + 384) = √2884 ≈ 53.69
The two possible values for t are:
t = (-(-50) ± 53.69) / (2 × 16) = (50 ± 53.69) / 32
We can discard the negative value since time cannot be negative, leaving us with:
t = (50 + 53.69) / 32 ≈ 3.35 seconds
Therefore, the correct answer is not listed in the provided options.
it hits the ground when the height is zero, right?
So just solve
6 + 95 sin25° t - 16t^2 = 0
Parametric Functions Test Part 1 Answers:
1.D 2. B and E 3. A B and D 4. B 5. B 6. A 7-9 Essays 10. C 11. B 12. a and E 13. B and C 14. A 15. C 16. A 17. B 18. A 19. B 20. A 21. A I just took the test these are 100% correct