Question

A ball is thrown into the air from the top of a building. The height, h(t), of the ball above the ground t seconds after it is thrown can be modeled by h(t) = -16t2 + 64t + 80. How many seconds after being thrown will the ball hit the ground?

Answer 1

Answer: 5 seconds

Step-by-step explanation:

Given : A ball is thrown into the air from the top of a building.

The height, h(t), of the ball above the ground t seconds after it is thrown can be modeled by $h(t) = -16t^2 + 64t + 80$ .

When ball reaches the ground , then its height from ground become zero.

i.e. $h(t) = -16t^2 + 64t + 80=0$

Divide equation by 16 , we get

$-t^2 + 4t + 5=0\\\\\Rightarrow\ t^2-4t-5=0\\\\\Rightarrow\ t^2+t-5t-5=0\\\\\Rightarrow\ t(t+1)-5(t+1)=0\\\\\RIghtarrow\ (t+1)(t-5)=0 \\\\\Rightarrow\ t= -1 , 5$

Since time cannot be negative , therefore t= 5

Hence, the ball will take 5 seconds ( after being thrown) to hit the ground.