Cannonball problem

Question

A cannonball is fired with an initial upward velocity of 100 m/s. The cannonball reaches its highest point halfway between the time it was fired (t=0) and the time it hits the ground.

When is it at its highest point.

How high is it at this point.

Answer 1

Let y = height 

 

y = y_0 + v_0y * t - (1/2) g t^2

 

y_0 = initial height = 0

 

v_0y = initial vertical speed = 100 m/s

 

Simplifying,

 

y = v_0y * t - (1/2) g t^2 

 

Set this equal to zero to find when it hits the ground again:

 

v_0y * t - (1/2) g t^2 = 0

 

t [v_0y -(1/2) g t] = 0

 

This has two solutions:  t = 0, and t = 2 * v_0y / g.  The value of 2 * v_0y / g is the time it reaches the ground again.  You want half of this time, so the time at maximum height is:

 

t = v_0y / g = (100 m/s) / (9.8 m/s^2) = 10.2 s

 

Now substitute this time back into y = v_0y * t - (1/2) g t^2 to get y_max:

 

y_max = (100 m/s)(10.2 s) - (1/2)(9.8 m/s^2) (10.2 s)^2 = 510.2 m