A ball is thrown from the top of a building with an initial velocity of 22.3 m/s

Question

A ball is thrown from the top of a building with an initial velocity of 22.3 m/s straight upward, at an initial height of 59.6 m above the ground. The ball just misses the edge of the roof on its way down, as shown in the figure.

(a) Determine the time needed for the ball to reach its maximum height.
s

(b) Determine the maximum height.
m

(c) Determine the time needed for the ball to return to the height from which it was thrown, and the velocity of the ball at that instant.

(d) Determine the time needed for the ball to reach the ground.
s

(e) Determine the velocity and position of the ball at t = 5.46 s.

m

A projectile is launched straight up at 52 m/s from a height of 83.5 m, at the edge of a sheer cliff. The projectile falls, just missing the cliff and hitting the ground below.

(a) Find the maximum height of the projectile above the point of firing.
m

(b) Find the time it takes to hit the ground at the base of the cliff.
s

(c) Find its velocity at impact.
m/s

Explanation / Answer

a)

time taken by ball to reach maximum height = v0/g

time taken by ball to reach maximum height = 22.3/9.8

time taken by ball to reach maximum height = 2.28 s

the time taken by ball to reach maximum height is 2.29 s

b)

maximum height = 59.6 + v0^2/(2*g)

maximum height = 59.6 + 22.3^2/(2 * 9.8)

maximum height = 85 m

the maximum height is 85 m
c)

time taken to reach to it's initial position = 2 * v0/g

time taken to reach to it's initial position = 2 * 2.28

time taken to reach to it's initial position = 4.56 s

the time taken to reach to it's initial position is 4.56 s

as there is no air resistance,

the velocuty at this time is 22.3 m/s downwards

d)

using second equation of motion

y = u*t + 0.5 at^2

-59.6 = 22.3 * t - 0.5 9.8 * t^2

solving for t

t= 6.44 s

the time taken to reach ground is 6.44 s

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