Question

A ball is thrown into the air. Its height (in feet) t seconds later is given by h(t)=64t−16t^2 a) What is the height of the ball 1 second after it has been thrown? b) After how many seconds does the ball hit the ground? c) At what time(s) is the ball 30 feet above the ground?

Answer 1

Answer:

A) 48 feet

B) 4 seconds

C) 0.54 seconds and 3.46 seconds

Step-by-step explanation:

So we have the function:

$h(t)=64t-16t^2$

Which represents the height h(t) of the ball after t seconds.

A)

To find the height of the ball 1 second after it has been thrown, substitute 1 for t. Thus:

$h(1)=64(1)-16(1)^2$

Simplify:

$h(1)=64-16$

Subtract:

$h(1)=48$

So, after 1 second, the height of the ball is 48 feet above the ground.

B)

To find out how many seconds after being thrown does the ball hit the ground, we'll set h(t) to 0. 0 means that it will be 0 feet above ground level: in other words, on the ground. So:

$0=64t-16t^2$

Solve for t. Factor out a 16t:

$0=16t(4-t)$

Zero Product Property:

$16t=0\text{ or } 4-t=0$

Solve for t:

$t=0\text{ or } t=4$

We'll ignore the first zero. So, the ball lands after 4 seconds of being thrown into the air.

C)

To find out at what time(s) the ball is 30 feet above the ground, set h(t) to 30 and solve for t:

$30=64t-16t^2$

First, divide everything by 2 to simplify things:

$15=32t-8t^2$

Subtract 15 from both sides:

$-8t^2+32t-15=0$

Now, to find the zeros, we can use the quadratic formula. a is -8, b is 32, and c is -15. Thus:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substitute:

$x=\frac{-32\pm\sqrt{32^2-4(-8)(-15)}}{2(-8)}$

Simplify:

$x=\frac{32\pm\sqrt{544}}{16}$

Use a calculator. Approximate:

$x\approx0.5422\text{ or } x\approx 3.4577$

So, the ball is 30 feet above the ground after 0.54 seconds and 3.46 seconds.

And we're done :)