Height of A Baseball

Question

A baseball is propelled upward with a speed of 64 feet per second from a height of 5 feet. For how many seconds is the ball higher than 53 feet?

Answer 1

The height equation is

h(t)=-16t²+64t+5

If graphed, this is a down-facing parabola with a vertex at t = -b/(2a) or t = -64/(2*-16) = -64/-32 = 2

So when t= 2 seconds, the baseball has reached its height. And that maximum height is h(2) 69 feet.

So the ball will pass through the height of 53 feet on its way up to its max height of 69 feet. And it will pass back through a height of 53 feet on its way back down. So if we figure out the time that the height is 53 feet on its way up, and the time when the height is 53 feet on its way down, we can find out how long it spent above 53 feet.

So set 53 = -16t²+64t+5

move the 53 over and get

0 =16t²+64t - 48, which factors into

-16(t²-4t+ 3) = -16(t - 1)(t - 3) = 0

So t=1 when it passes through 53 feet high on the way up.

And t = 3 when it passes through 53 feet on the way down.

Between 1 second and 3 seconds there are 2 seconds that the ball spent above 53 feet.