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 [tex]t=3 seg[/tex]
[tex]h(3)=-16(3)^{2} +63(3)+4[/tex]
[tex]h(3)=-144+189+4\\h(3)=-144+193\\h(3)=49 feet[/tex]
B. What is the maximum height of the ball? round to the nearest foot
[tex]h(t)=-16(t)^{2} +63(t)+4\\h'(t)=-32(t) +63\\[/tex]
then
[tex]-32(t)+63=0\\t=\frac{-63}{-32}\\t=1.969seg[/tex]
For [tex]t=1.969seg[/tex]
[tex]h(1.969)=-16(1.969)^{2}+63(1.969)+4\\h(1.969)= 66.016\\h(1.969)=66 feet[/tex]
C. When will the ball hit the ground?
The ball will hit the ground when [tex]h(t)=0[/tex]
so, [tex]-16t^{2} +63(t)+4=0[/tex]
Using the quadratic equation
[tex]t=4seg[/tex]
