Question

If a projectile (such as cannonball) is fired into the air with an initial velocity at an angle of elevation , then the height of the projectile at time t is given by . If the cannonball is shot with an initial velocity of 600 feet per second with an angle of elevation at , determine the equation of , in exact form. What is the height of the canon after 2 seconds?

Answer 1

Answer: Hi, first, the cannon only gives the cannonball the initial velocity, and when the cannonball is in the air, the only force acting on the ball is the gravitational force.

First, let's compute the initial velocity, if the ball is fired with angle A (measured from the ground, or +x in this case) and velocity V0. then the vector of the velocity is (cos(A)*V0, Sin(A)*V0)

now start describing all the equations.

Acceleration, we know that an object in the air will fall with acceleration g = 9.8 m/s.

then a(t) = (0, -g)

Velocity: integrating the acceleration over the time, we obtain v(t) = (0,-g*t) +C

where C is a integration constant, equal to the initial velocity. Then v(t) = (cos(A)*V0, Sin(A)*V0 - g*t)

Position; For the position we need to integrate again over time, then:

p(t)= (cos(A)*V0*t, Sin(A)*V0*t - $g*\frac{t^{2} }{2}$) + K.

where again K is an integration constant, in this case the initial position, that  write it as (X0,Y0).

The height of the cannonball after 2 seconds is the y component valued in t=2

height = Y0 + Sin(A)*V0*2 - $g*\frac{2^{2} }{2}$.

where you can put the angle A and the initial velocity V0 to obtain the height.