h(t)=16t^2+30t

Question

A scoop in field hockey is a pass that propels the ball from the ground into the air. Suppose a player makes a scoop with an upward velocity of 30 ft/sec. The function h(t)=16t^2+30t models the height H in feet and time t in seconds.
A.) What is the height of the ball after 1.5 seconds
B.) Will the ball ever reach the height of 16 feet?
C.) What is the maximun height of the ball, and after how many seconds it is achieved?

Answer 1

This is a physics problem...

 

The function given is the position function. 

Can you first please check if you have written the equation right?? Typically it is -16t²+30t.  The -16 is the force of gravity in feet per second. 

 

If -16t²+30t is correct in part A you will plug in 1.5 into t.  This will give you -16(2.25)+45 =9

 

For part B you can  plug in 16 into y so you get 16 = -16t²+30t.  You will then take the 16 over to get 0 = -16t²+30t-16.

 

After that you can use quadratic equation to solve it.  If you do that you will get a negative square root so it will never reach 16 feet. 

 

Finally part C is what is its maximum height.  In this you can use x=-b/2a which will give you 30/32 or .9375.  You will take this point and plug it back into t and see what you get for y which is the height.  After I did it I got 14.0625 feet.  Thanks and good luck!!!!!