A ball is thrown up from the top of a 64 foot high building. The ball is 96 feet of the ground after 1 second and 84 feet off the ground after 2.5 second.

Question

A= how long after the ball is thrown will it reach its maximum height ?

B= what is the maximum height the ball reaches?

C= how long after the ball is thrown will it hit the ground?

Answer 1

typical position function is : x(t) = -16* t^2 + Vo * t 0+ 64 where Vo is the initial velocity.

We have (1,96), (2.5, 84) and (0,64) <--- initial position at time zer0

x(t) = At^2 + Bt + C

(0,64) ---> X(0) = 64 ---> C=64 as expected

(1,96) ---> X(1) = 96 ----> A+B+64 = 96

A+B = 32

(2.5, 84) ---> X(2.5) = 84 ---> 6.25A + 2.5B + 64 = 84

6.25A + 2.5B = 20

6.25A + 2.5B = 20

A + B = 32

-2.5 * bottom equation + top equation:

3.75A = -60

A = -16

as expected

Per the bottom equation, B = 48

x(t) = -16t^2 +48t + 64