A punter on Westview High School's 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 foot?

Question

anonymous · Answer

We can solve this in a couple of different ways:

1) Using the Equations of projectile motion.

We know that:
$$V_{f}=V_{i}-g \times t $$
We also know that at the ball's maximum height the velocity will be zero. (i.e. The ball has to stop moving upwards before it can start falling back down. For just an instant it's not moving at all.)

i.e. $$V=0;h=\max$$ 

Plugging these known values into out first equation and using the known value for the acceleration due to gravity (NOTE: In this question they seem to be using 32 feet per second squared. A more accurate value is about 32.17 feet per second squared, but we;'ll just use what they gave us.) 

So:
$$0ft/s=45ft/s-(32ft/s ^{2} \times t)$$
or:
$$t= \frac{ 45ft/s }{ 32ft/s ^{2} } = 1.406 seconds$$

We now know that it takes 1.406 seconds for the ball to reach its maximum height before it starts falling back down. We can use this fact to find the maximum height.

We know that:

$$h=V \times t - \frac{ g \times t^{2} }{ 2 } $$

Plugging in everything we know:
$$h=45ft/s \times 1.406 seconds - \frac{ 32ft/s^{2} \times 1.406 seconds^{2} }{ 2 } $$
and we get

$$h_{\max} = 31.64 feet$$

2) Using derivative calculus.  

We have:
$$h=-16t^2+45t+5$$

We also know that velocity is the first derivative of position (i,.e. Velocity is a change in position over time.)

$$V = \frac{ dh }{ dt }$$

So, taking the first derivative of out position function we can get a new function for the balls velocity:

$$V=\frac{ dh }{ dt } = -32t+45$$

(Notice that this is the same formula for velocity that we got with method 1)

Similar to method 1 we want to find the point where the velocity is equal to zero. i.e. the point where the first derivative of the position function is equal to zero. Graphically this is either the maximum or minimum point on the graph of the function. We want the maximum.

So:

$$h _{\max}=\frac{ dh }{ dt } @ 0$$ 
$$0=-32t+45$$

as before:

$$t=1.406seconds $$

and 

$$h_{\max} = 31.64 feet$$