Question

A cannonball is fired across a flat field at an angle of 43 degrees with an initial speed 32 m/s and height of 12 m.

1)Write a set parametric equations for the motion of the cannon ball
2)Determine how long the cannon ball was in the air
3)Determine how far the cannon ball traveled in the air
4) Determine when the cannon ball reaches its maximum height
5) Determine the maximum height reached by the cannon ball

Answer 1

1)
`x= v_(0x) t = 23.4 t\\y=y_0 + v_(0y)t-(1)/(2)gt^2 = 12+21.8t -4.9t^2`

The initial data of the projectile are:

`y_0 = 12 m` is the initial height

`v_0 = 32 m/s` is the initial speed of the projectile, so its components along the x- and y- directions are

`v_(0x) = v_0 cos \theta = (32 m/s)(cos 43^(\circ))=23.4 m/s\\v_(0y) = v_0 sin \theta = (32 m/s)(sin 43^(\circ))=21.8 m/s`

The motion of the cannonball along the x-direction is a uniform motion with constant speed, while on the y-direction it is an uniformly accelerated motion with constant acceleration g=9.8 m/s^2 downward. So, the two equations of motion of the projectile along the two directions are:

`x= v_(0x) t = 23.4 t\\y=y_0 + v_(0y)t-(1)/(2)gt^2 = 12+21.8t -4.9t^2`

2) 4.94 s

To determine how long the cannon ball was in the air, we need to find the time t at which the cannonball hits the ground, so the time t at which y(t)=0:

`0=12+21.8t-4.9 t^2`

Solving the equation with the formula, we have:

`t_(1,2)=(-21.8\pm √((21.8)^2-4(-4.9)(12)))/(2(-4.9))`

which has two solutions:

t = -0.50 s

t = 4.94 s

Discarding the first solution which is a negative time so it has no physical meaning, the correct solution is

t = 4.94 s

3) 115.6 m

To determine how far the cannonball travelled, we need to find the value of the horizontal position x(t) when the ball hits the ground, at t=4.94 s. Substituting this value into the equation of motion along x, we find:

`x=v_(0x)t=(23.4 m/s)(4.94 s)=115.6 m`

4) 2.22 s

The cannonball reaches its maximum height when the vertical velocity becaomes zero.

The vertical velocity at time t is given by

`v_y(t)= v_(0y) -gt`

where

g = 9.8 m/s^2 is the acceleration due to gravity

Substutiting
`v_y(t)=0` and solving for t, we find

`t=(v_(0y))/(g)=(21.8 m/s)/(9.8 m/s^2)=2.22 s`

5) 36.2 m

The maximum height reached by the cannon is equal to the vertical postion y(t) when the vertical velocity is zero, so when t=2.22 s. Substituting this value into the equation of the vertical motion, we find:

`y(t)=y_0 + v_(0y)t-(1)/(2)gt^2=12+(21.8)(2.22)-(4.9)(2.22)^2=36.2 m`