Question

A ball is thrown from the top of a building with a velocity of 10 m/s and returns with a velocity of 30 m/s to the ground. Find the height of the building.​

Answer 1

Explanation:

Let’s review the 4 basic kinematic equations of motion for constant acceleration (suggest you commit these to memory):

s = ut + ½at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, we know u = 30m/s, a = -9.81m/s^2, and we want to know:

a) the max height of the ball, so we use equation (2)

0 = 30^2 – 2(9.81)s then s = 900/19.62 = 45.872m

b) the time to reach max height, so we use equation (4)

45.872 = (30 + 0)t/2 then t = 45.872/15 = 3.058 seconds

c) the height of the building if v = 90m/s, so we use equation (2)

90^2 = 30^2 – 2(9.81)s then s = (8100 – 900)/19.62 = 366.972m say 367m