Question

Someone please help with parts a b c and d. I will mark brainlest!!!

Answer 1

Answer:

A. 49 feet

B. 66 feet (round to the nearest foot)

C. 4 seg

Step-by-step explanation:

A. What is the height of the ball after 3 seconds?

For $t=3 seg$

$h(3)=-16(3)^{2} +63(3)+4$

$h(3)=-144+189+4\\h(3)=-144+193\\h(3)=49 feet$

B. What is the maximum height of the ball? round to the nearest foot

$h(t)=-16(t)^{2} +63(t)+4\\h'(t)=-32(t) +63$

then

$-32(t)+63=0\\t=\frac{-63}{-32}\\t=1.969seg$

For $t=1.969seg$

$h(1.969)=-16(1.969)^{2}+63(1.969)+4\\h(1.969)= 66.016\\h(1.969)=66 feet$

C. When will the ball hit the ground?

The ball will hit the ground when $h(t)=0$

so, $-16t^{2} +63(t)+4=0$

Using the quadratic equation

$t=4seg$