A rescue helicopter drops a crate of supplies from a height of 160 feet. After t seconds, the crate is 160 - 16t^2 feet above the ground.

a) Find the velcocity of the crate at t=1
b) With what velocity does the crate strike the ground?

for a do i just find the derivative and plug 1 in for t?

h=160-16t^2

velocity is dh/dt, so yes on a)
Now, for b, find t when h=0 Put that t into the dh/dt equation to find velocity.

Yes, to find the velocity of the crate at t = 1, you would need to find the derivative of the height function and then substitute t = 1 into that derivative.

Let's first find the derivative of the height function 160 - 16t^2 with respect to time t. The derivative gives us the rate of change of height with respect to time, which is velocity.

So, the derivative of 160 - 16t^2 would be:

d/dt (160 - 16t^2) = 0 - 32t = -32t

Now, we can substitute t = 1 into this derivative:

Velocity at t = 1: v(1) = -32(1) = -32 feet per second

Therefore, the velocity of the crate at t = 1 is -32 feet per second.

To find the velocity with which the crate strikes the ground, we need to find the time t when the height is 0. This is because the height of the crate above the ground is 0 when it strikes the ground.

Setting the height function to 0 and solving for t:

160 - 16t^2 = 0

Dividing both sides by 16:

10 - t^2 = 0

Rearranging:

t^2 = 10

Taking the square root of both sides:

t = ±√10

Since we are interested in the positive time t, we can discard the negative solution.

Therefore, the crate strikes the ground at t = √10 seconds.

To find the velocity with which the crate strikes the ground, we can substitute t = √10 into the derivative we found earlier:

Velocity at t = √10: v(√10) = -32(√10) = -32√10 feet per second

Hence, the crate strikes the ground with a velocity of -32√10 feet per second.