You drop a ball off a roof and its position in feet s at tim

You drop a ball off a roof and its position (in feet), s, at time t (in seconds) is given by s(t)= -16t²+100

How high was the ball when you initially dropped it?

How long until the ball hits the ground?

Estimate the instantaneous velocity of the ball when t=2 using smaller intervals. Why is this number negative?

Solution

a) The ball is released at time t = 0s. So substitute t = 0 in the expression: s(t) = -16t² + 100

this will give: s(t) = -16(0)² + 100 = 100

so the ball was at a height of 100 feet from the ground.

b) at time T , the ball hits, the ground. So the height s(t) = 0 ft

therefore substitute this in the expression for s(t) to get:

0 = -16t² + 100

=> 16t² = 100

t = (100/16)^1/2 = 2.5 seconds.

c) for time t = 2s, use the equation v = u + at

here the initial velocity u = 0 m/s

and a = - 9.8 m/s² [since it is acting downwards].

therefore v = 0 - 9.8 x 2 = -19.6 m/s = - 64.304 ft/s

This value of velocity is negative because we conventionally choose upward direction to be positive. Therefore the acceleration due to gravity is in the opposite direction [downwards].