Question

A ball is launched into the sky at 54. 4 ft./s from a 268.8 m tall building. The equation for the ball’s height, h, at time t second’s is h= -3.2t^2 + 54.4t+ 268.8. When will the ball strike the ground?

Answer 1

Answer:

t = 21

Step-by-step explanation:

The ball strikes the groujd when h = 0.

$-3.2t^2 + 54.4t+268.8=0 \\ \\ 32t^2 - 544t-2688=0 \\ \\ t^2 - 17t-84=0 \\ \\ (t-21)(t+4)=0 \\ \\ t=-4, 21$

Since time must be positive, t = 21.