well i have this homework question that says:

A rock is dropped from the top of a 300-ft cliff. It's velocity at time t seconds is v(t0= -32 t feet per second.

a.) Find the height of the rock above the ground at time t.

b.) How long will the rock take to reach the ground?

c.) What will be the velocity when it hits the ground?

Assume that the velocity at t=0 is zer, sinc it is "dropped" and not thrown.

The acceleration rate is a = -g = -32 ft/s^2
a) height above ground = 300 -(a/2) t^2
= 300 - 16 t^2
b) solve 300 - 16 t^2 = 0
t = sqrt (300/16) = 4.33 s

c) Solve V = -32 t when t = 4.33 s

To solve the given problem, we'll need to use the equations of motion for a free-falling object. Let's break down the problem into three parts and solve each part separately.

a.) Finding the height of the rock above the ground at time t:

The equation that describes the position of the object in terms of time is given by:
s(t) = s₀ + v₀t + (1/2)at²

In this equation:
- s(t) is the position (or height) at time t.
- s₀ is the initial position (or height).
- v₀ is the initial velocity.
- t is the time.
- a is the acceleration due to gravity, which is -32 ft/s² (negative because it acts downward).

Given that the rock is dropped from the top of a 300-ft cliff and the initial velocity is 0 ft/s, we can substitute these values into the equation:
s(t) = 300 + 0t + (1/2)(-32)t²

Simplifying the equation gives us:
s(t) = 300 - 16t²

Therefore, the height of the rock above the ground at time t is given by the equation s(t) = 300 - 16t².

b.) How long will the rock take to reach the ground:

To find the time it takes for the rock to reach the ground, we need to solve the equation s(t) = 0 (since the rock is at the ground level).

Setting s(t) = 0 in the equation s(t) = 300 - 16t²:
300 - 16t² = 0

Simplifying the equation gives us:
16t² = 300

Dividing both sides by 16:
t² = 18.75

Taking the square root of both sides:
t = ±√18.75

Since we're dealing with time, the negative root doesn't make sense in this context. Therefore:
t = √18.75

Calculating the value of √18.75 gives us:
t ≈ 4.33 seconds

So, the rock will take approximately 4.33 seconds to reach the ground.

c.) What will be the velocity when it hits the ground:

The velocity of the rock when it hits the ground can be found using the equation for velocity:
v(t) = v₀ + at

Given that v₀ = 0 ft/s (since the rock is dropped from rest) and a = -32 ft/s², we can substitute these values into the equation:
v(t) = 0 + (-32)t

Simplifying the equation gives us:
v(t) = -32t

Therefore, the velocity of the rock when it hits the ground is v(t) = -32t ft/s.