Math Component of Cannon

h = -16t²+v₀t+h₀

h= height at a certain time

t= time the cannonball is in the air

v0= initial velocity

h0= starting height

This equation can be used to determine how long an object is in the air, its maximum height, or its initial velocity. In this case, the variables can be used to solve the following problems:

A cannonball is shot upward from the upper deck of a fort with an initial velocity of 192 feet per second.  The deck is 32 feet above the ground.

 1. How high does the cannonball go? 608 feet
2. How long is the cannonball in the air? 12.15 seconds

h= -16t²+192t+32

1.    Vertex (t)=_-b         v₀ or b= 192 f/s      a= -16

            2a

t= -(192)/ 2(-16)= 6

h= -16(6)²+192(6)+32

h= -576+1152+32

h=608 ft.

2.     The quadratic equation can be used to plug in the numbers.

 

          ≈12.15 seconds

         

Angle Justification

Using my knowledge of geometric angles and their functions, I began to narrow down my search for the "perfect" angle at which to launch an object. A 90 degree angle would have the effect of shooting an object straight up. A 0 (or 180) degree angle would roll an object straight along the ground with very little gain. Therefore, the perfect angle seemed to be an angle exactly halfway in between each of these measurements. A 45 degree angle would get the maximum distance while maintaining the greatest height possible without effect on the horizontal distance.