Question

A ball is thrown into the air with an upward velocity of 80 feet per second. The function h = -16t^2 + 80t models the height h, in feet, of the ball at time t, in seconds. When will the ball reach the ground? after 3 seconds after 4 seconds after 5 seconds after 6 seconds

Answer 1

Answer:

5 seconds

Step-by-step explanation:

The given height h, in feet, of the ball at time t, in seconds is defined by the function

$h=-16t^2+80t$

The height of the ball is 0 if it reached on the ground.

Substitute h=0 in the given function.

$0=-16t^2+80t$

$0=-16t(t-5)$

Using 0 product property, we get

$-16t=0\Rightarrow t=0$

$(t-5)\Rightarrow t=5$

It means ball will reach the ground at t=0(initial condition) and t=5.

Therefore, the ball will reach the ground after 5 seconds.

Answer 2

The ball hits the ground when the height  h(t) = 0 , so

-16t^2 + 80t = 0

-16t(t - 5) = 0

t- 5 = 0  or -16t^2 = 0 (  here t = 0 which corresponds to when ball is thrown)

t - 5 = 0 gives:-

t = 5 seconds  (answer)