Question

The height, h(t), in feet of an object thrown into the air with an initial upward velocity of 63 feet per second is given by the formula h(t) = -16t2 + 63t, where t is the time in seconds. What is the height, in feet, of the object after 3 seconds?

Answer 1

Answer:

The height of the object after 3 seconds is 45 feets.

Step-by-step explanation:

The height of an object thrown into the air with an initial upward velocity of 63 feet per second is given as

$h(t)=-16t^2+63t$

Where, h(t) is height of the object in feet and t is the time in seconds.

We have to find the height of the object after 3 seconds. So, substitute t=3.

$h(3)=-16(3)^2+63(3)$

$h(3)=-16(9)+189$

$h(3)=-144+189$

$h(3)=45$

Therefore height of the object after 3 seconds is 45 feets.