Question

suppose a ball thrown into the air has its height (in feet) given by the function ( h(t) = 15 + 160t - 16t^2 ), where ( t ) is in seconds after the ball is thrown.

a. find ( h(1) ) and explain what it means.

b. find the height of the ball 5 seconds after it is thrown.

c. test other values of ( h(t) ) to decide if the ball eventually falls. when does the ball stop climbing?

Explanation:

Step1: Calculate h(1)

Substitute \(t=1\) into \(h(t)\):
\(h(1)=15+160(1)-16(1)^2=15+160-16=159\)
This is the height (in feet) of the ball 1 second after being thrown.

Step2: Calculate h(5)

Substitute \(t=5\) into \(h(t)\):
\(h(5)=15+160(5)-16(5)^2=15+800-400=415\)

Step3: Find when climbing stops

The ball stops climbing at the vertex of the quadratic function. For \(h(t)=at^2+bt+c\), the vertex time is \(t=-\frac{b}{2a}\). Here \(a=-16\), \(b=160\):
\(t=-\frac{160}{2(-16)}=\frac{160}{32}=5\)

Step4: Test values for falling

Test \(t=6\) (after \(t=5\)):
\(h(6)=15+160(6)-16(6)^2=15+960-576=399\)
Since \(h(6)<h(5)\), the ball falls after \(t=5\).

Answer:

a. \(h(1)=159\); this is the height of the ball (159 feet) 1 second after it is thrown.
b. \(h(5)=415\) feet
c. The ball stops climbing at \(t=5\) seconds. Testing \(t=6\) gives \(h(6)=399\) feet, which is less than \(h(5)\), so the ball eventually falls.