Question

A ball is thrown into the air from a height of 4 feet at time t = 0. The function that models this situation is h(t) = -16t2 + 63t + 4, where t is measured in seconds and h is the height in feet.
a) What is the height of the ball after 3 seconds?
b) What is the maximum height of the ball? Round to the nearest foot.
c) When will the ball hit the ground?
d) What domain makes sense for the function?

Answer 1

a) To find the height of the ball after 3 seconds we substitute into the function t=3:

$$h\left( 3 \right)=-16\cdot {{3}^{2}}\text{ }+\text{ }63\cdot 3\text{ }+\text{ }4=-144+189+4=49\,\text{feet}$$

So h(3)=49 feet.

b) To find the maximum height of the ball, we first find the derivative of the function that defines this height.

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

Then equate this to 0

$$-32t+63=0\,\,\,=>\,\,\,t=\frac{63}{32}\approx 1.97\,\text{s}$$

We also see that h''(t) = -32, which is negative, so at t = 1.97 seconds we have a maximum value of h.

At time equal to 1.97 seconds the height of the ball is

$$h\left( 1.97 \right)=-16\cdot {{\left( 1.97 \right)}^{2}}\text{ }+\text{ }63\cdot 1.97\text{ }+\text{ }4=-62.1+124.1+4=66\,\text{feet}$$

c) To find time when the ball hit the ground we equate h(t) to zero

$$h\left( t \right)=-16{{t}^{2}}+63t+4=0$$

and solve this for t

$$t=\frac{-63\pm \sqrt{{{63}^{2}}-4\cdot \left( -16 \right)\cdot 4}}{2\cdot \left( -16 \right)}$$

$$t=\frac{-63\pm \sqrt{3969+256}}{-32}$$

$$t=\frac{-63\pm \sqrt{4225}}{-32}$$

$$t=\frac{-63\pm 65}{-32}$$

So time when the ball hit the ground is t₁=4 seconds (second solution t₂=-1/16<0 is wrong).

d) The domain for the function is the set of all possible t-values which will make the function "work"

So we have the domain that makes sense for this function

$$0\le t\le 4$$