A ball is thrown vertically upward with an initial velocity of 80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is
s(t) = 80t - 16t^2.
a) At what time t will the ball strike the ground?
b) For what time t is the ball more than 96 feet above the ground?

Question

A ball is thrown vertically upward with an initial velocity of 80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is 
s(t) = 80t - 16t^2.
a) At what time t will the ball strike the ground?
b) For what time t is the ball more than 96 feet above the ground?

snowfire · Answer

For part a, you simply need to find the time it takes to reach the peak (derivative=0 at the peak) and double it to get your time to land. And for part b, you want to find the two points where the position function equals 96, which you will probably find using the quadratic equation.

snowfire · Answer

Oh and also for the second part, you want to find the difference in time between those two points.

anonymous · Answer

One method for the first part:
Find the derivative of the function for distance that you have, giving you the function for velocity. Equate this to zero and you get the time 2.5s.
$$v(t) = 80 -32t$$
Double this and you have 5s.

The other method:
Simple equate the height function you have to 0, giving you 5.

For the time when the object is above 96ft, simply equate the height function to 96. Subtract the two points you'll get by solving the quadratic function.

jack1 · Answer

"For part a, you simply need to find the time it takes to reach the peak (derivative=0 at the peak) ***and double it to get your time to land***."
@Lulu31 just an FYI, this doubling method only works if it;s thrown from ground level or landing at the same heigth it's thrown from

snowfire · Answer

Jack is correct, forgot to mention that ^^"

snowfire · Answer

Does any of this make sense @Lulu31 ?

anonymous · Answer

Yes thank you so much guys!!