Question

Paige launched a ball using a catapult she built. The height of the ball (in meters above the ground) ttt seconds after launch is modeled by h(t)=-5t^2+40th(t)=−5t 2 +40th, left parenthesis, t, right parenthesis, equals, minus, 5, t, squared, plus, 40, t Paige wants to know when the ball will hit the ground. 1) Rewrite the function in a different form (factored or vertex) where the answer appears as a number in the equation. h(t)=h(t)=h, left parenthesis, t, right parenthesis, equals 2) How many seconds after launch does the ball hit the ground? seconds

Answer 1

Answer:

1) Factor form : $h(t)=-5t(t-8)$

2) 8 second after launch.

Step-by-step explanation:

The height of the ball (in meters above the ground) t seconds after launch is modeled by

$h(t)=-5t^2+40t$

To find the time when ball hit the ground, we need to find the factor form of the given function.

$h(t)=-5t(t-8)$

When ball hi the ground, then height of the ball from the ground is 0.

$h(t)=0$

$-5t(t-8)=0$

Using zero product property, we get

$-5t=0\Rightarrow t=0$

$t-8=0\Rightarrow t=8$

Ball hit the ground at t=0 and t=8. It means ball hit the ground in starting and 8 second after launch.

Answer 2

Answer:

-5t(t - 8) and it hit 8 seconds after launch.