A ball is thrown off the top of a building with a velocity v at an angle ? with

Question

A ball is thrown off the top of a building with a velocity v at an angle ? with respect to the horizontal.
If the building is 50.0 m tall, the initial horizontal velocity vox=15.0 m/s and the initial vertical velocity
voy=20.0 m/s.

A) Find the ball's maximum height above the ground   (answer: 70.4 m)
B) Find the time it takes to reach the maximum height (answer: 2.04 s)

C) Find the total time of flight until the ball hits the ground
D) Find the horizontal range

*** I got A & B correct, but i keep getting C and D wrong

A ball is thrown off the top of a building with a velocity v at an angle ? with respect to the horizontal. If the building is 50.0 m tall, the initial horizontal velocity vox=15.0 m/s and the initial vertical velocity voy=20.0 m/s.

Explanation / Answer

angle ? = tan^-1(vox / voy) = tan^-1(12.0 m / 12.0 m) = 45 degrees

let s = displacement/distance of the ball
vfy = final vertical velocity of the ball
vfx = final horizaontal velocity of the ball

For the upward motion of the projectile:

a) at the maximum height:

s = [(vfy)^2 - (voy)^2] / 2(-g)
s = [0 - (12.0 m/s)^2] / 2(-9.8 m/s^2)
s = 7.35 meters

maximum height above the ground = 50.0 meters + 7.35 meters
maximum height above the ground = 57.35 meters

b) at the maximum height:

-g = (vfy - voy)/t
so therefore,
t = (0 - 12 m/s)/-9.8 m/s^2
t = 1.22 seconds (time it took to reach the maximum height)

c) for the downward motion of the projectile (from the maximum height until it reaches the ground)

s = (voy)(t) + 1/2(g)(t^2)
57.35 m = (0 m/s)(t) + 1/2(9.8 m/s^2)(t^2)
t = 3.42 seconds

total time of flight = 3.42 seconds + 1.22 seconds = 4.64 seconds

d) for the horizontal range of the projectile:

s = (vox)(t) = (12.0 m/s)( 4.64 sec.) = 55.68 meters

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