Question

A ball is thrown upward with an initial velocity of 16 ft/sec from a height of 32 ft above the ground. The height, in feet, of the ball t sec after it is thrown can be represented by s=-16t^2+16t+32. How long will it take for the ball to reach the ground?

Answer 1

Answer:  2 seconds.

Step-by-step explanation:

Given : A ball is thrown upward with an initial velocity of 16 ft/sec from a height of 32 ft above the ground.

The height, in feet, of the ball t sec after it is thrown can be represented by

$s=-16t^2+16t+32.$

When ball reaches the ground , s= 0.

$\Rightarrow -16t^2+16t+32=0$

Divide both sides by 16 , we get

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

But time cannot be negative , so t= 2.

Hence , it will take 2 seconds for the ball to reach the ground.