Question

An object is dropped from a tower, 183 ft above the ground. the height above ground t sec into the fall is s=183-16t^2

What is the object’s velocity, speed and acceleration at time t?

About how long does it take the object to hit the ground?

What is the object’s velocity at the moment of impact?

Answer 1

Explanation:

We are given the position function:

s(t) = 183 - 16
`t^(2)`

The velocity at t is the derivative of s(t), s'(t)

s'(t) = -32t

Given that the motion is in one direction, the speed would just be the magnitude of the velocity, or in this case the absolute value.

The acceleration at t is the second derivative of s(t), s''(t)

s''(t) = -32

The object will hit the ground when s(t) = 0

0 = 183 - 16
`t^(2)`

16
`t^(2)` = 183

`t^(2)` = 183/16

t =
`√(183/16)`

The objects velocity at the moment of impact is simply t we solved for earlier substituted into s'(t).

s'(sqrt(183/16)) = -32 * \sqrt{183/16}