If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t - 16t^2

What is the velocity of the ball when it is 96 ft above the ground on its way up?

Solve the equation

16t^2 -160t +96 = 0
There will be two roots. Take the lowest t value as the solution. That will be the answer when the ball is going up.

Note that the equation can be simplifed to
t^2 -10t + 16 = 0
which factors easily.

To find the velocity of the ball when it is 96 ft above the ground on its way up, we need to differentiate the height function with respect to time and then substitute the given height into the resulting expression.

First, let's differentiate the height function:
s(t) = 160t - 16t^2

To differentiate s(t), we can differentiate each term separately since they are both functions of t.

The derivative of 160t with respect to t is simply 160, since the derivative of t with respect to t is 1.

The derivative of -16t^2 with respect to t requires using the power rule for differentiation. According to the power rule, the derivative of t^n with respect to t is n * t^(n-1). Therefore, the derivative of -16t^2 with respect to t is -16 * 2 * t^(2-1), which simplifies to -32t.

Now let's substitute t = 96 into the derivative expression we obtained:
v = 160 - 32t

Substituting t = 96 into v gives:
v = 160 - 32(96)
v = 160 - 3072
v = -2912 ft/s

Therefore, the velocity of the ball when it is 96 ft above the ground on its way up is -2912 ft/s. Note that the negative sign indicates that the ball is moving upward.