Question

A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is 1. After how many seconds does the ball strike the ground?

Answer 1

Answer:

$t=6.96s$

Explanation:

From this exercise, our knowable variables are hight and initial velocity

$v_{oy}=96ft/s$

$y_{o}=112ft$

To find how much time does the ball strike the ground, we need to know that the final position of the ball is y=0ft

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}$

$0=112ft+(96ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}$

Solving for t using quadratic formula

$t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}$

$a=-\frac{1}{2} (32.2)\\b=96\\c=112$

$t=-0.999s$ or $t=6.96s$

Since time can't be negative the answer is t=6.96s