Question

A canon ball is fired with an initial velocity of 123.6 m/s at an angle of 32.9° above the horizontal. Calculate the...?

a. the range
b. the time the ball is in the air (total time)
c. the final vertical velocity of the ball

Answer 1

(a)

1420.4 m

(b)

21.15 s

(c)

103.8 m/s

Explanation:

(a)

For the horizontal range, assuming level ground, we can use:

`sf(d=(v^2sin2theta)/(g))`

`:.` `sf(d=(123.6^2xx0.9121)/(9.91)color(white)(x)m)`

`sf(d=1420.4color(white)(x)m)`

(b)

The horizontal component of velocity `sf(v_x)` is constant and is equal to `sf(vcostheta)`.

So we can say:

`sf(t=(d)/(vcostheta))`

`:.` `sf(t=(1420.4)/(123.6xx0.54317)color(white)(x)s)`

`sf(t=21.15color(white)(x)s)`

(c)

If we consider only the vertical component of the cannon ball's motion we can use the equation of motion:

`sf(v=u+at)`

At the cannon ball's maximum height we can say that `sf(u=0)` and, because of the symmetry of the flight path, we can say that the time to reach the ground will be `sf(t/2)`.

So we get:

`sf(v_y=0+("g"t)/2)`

`:.` `sf(v_y=9.81xx(21.15)/(2)=103.8color(white)(x)"m/s")`