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

1 Answer
Oct 29, 2017

(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")