Question

A punter on westveiw high schools football team kicks a ball straight up from 5 feet above the ground with an initial upward velocity of 45 feet per second. the height of the ball above ground after t seconds is given by the equation H + -16t^2 + 45t + 5, where h is the height of the ball in feet and t is the time in seconds since the punt. what is the maximum height of the ball, to the nearest feet?

A:37 feet
B:38 feet
C:39 feet
D:400 feet

Answer 1

The equation for the height, if graphed (h on y axis and t on x axis) will give a smooth curve that rises, reaches a maximum and falls. It might be good to sketch that.
What we are after is the maximum value of h that is reached. Now, note that at the maximum point of the graph, the gradient at that point is zero.
Our equation is h=-16t^2+45t+5
Calculate the gradient of the graph by differentiating this equation.
Gradient=-32t+45
As stated before, the gradient at highest point is zero so,
0=-32t+45
32t=45
t=45/32
So we have the time at which the highest point is reached. Plug in that value of t
h=(-16(45/32)^2)+(45*45/32)+5
Calculate the value of h and then add 5 to it (remember the ball was kicked while 5 feet off the ground)

Answer 2

h=36.6406, approx 37feet

Answer 3

As stated in my first post, I believe we should add 5 to 37 to get 42 feet. But then that is not in the choices. :)

Answer 4

Thanks. and i got it now . :)

Answer 5

You're welcome :)