Question

The height of a golf ball is given by h= =16t^2 + 48t, where t is in seconds and h is in feet.

a.) At what times is the golf ball on the ground?

Answer 1

\(\large h=-16t^2+48t\)

The golf ball will be on the ground just before it's hit, and also after it lands.
Another way to interpret this is, it will be on the ground when the `height` above the ground, \(\large h\), is 0.

So we'll let h=0, and solve for t.

Answer 2

\[\large 0=-16t^2+48t\]If we factor a \(\large t\) out of each term, we get,\[\large 0=t(-16t+48)\]

We can use the `Zero Factor Property` and set each factor equal to zero,\[\large 0=t \qquad \qquad \qquad 0=(-16t+48)\]

From here, solve for your t values! :)
The first solution should make sense.
When no time has passed, the ball is still sitting on the ground. That's at t=0.

Answer 3

K thanks so much! :)