Question

A ball is fired at a speed of 8 m/s at 20˚ above the horizontal from the top of a 5 m high wall. How far from the base of the ball will the ball hit the ground? Also calculate the angle below horizontal the velocity vector makes.

Answer 1

`sf(10color(white)(x)m)`

`sf(53.9^@)`

Explanation:

`sf((a))`

To get the time of flight we can use:

`sf(s=ut+1/2at^2)`

`sf(v_y=8sin20)`

Using the convention that "down is +ve" this becomes:

`sf(5=-8sin20t+1/2xx9.81xxt^2)`

This simplifies to:

`sf(4.905t^2-2.736t-5=0)`

Applying the quadratic formula:

`sf(t=(2.763+-sqrt(7.634-4xx4.905xx-5))/(9.81))`

Ignoring the -ve root this gives:

`sf(t=1.33color(white)(x)s)`

We can now get the range d from the horizontal component of velocity which is constant:

`sf(v_x=8cos20=d/1.33)`

`:.` `sf(d=8cos20xx1.33)`

`sf(d=7.517xx1.33)`

`sf(d=10color(white)(x)m)`

`sf((b))`

To get the angle to the horizontal which the ball strikes the ground we need the vertical component of its velocity when it lands.

We can use:

`sf(v=u+at)`

This becomes:

`sf(v_y=-8sin20+(9.81xx1.33))`

`sf(v_y=-2.736+13.034=10.3color(white)(x)"m/s")`

See fig (b):

We now know:

`sf(v_x=7.517color(white)(x)"m/s")`

`sf(v_y=10.3color(white)(x)"m/s")`

If we needed the resultant we could use Pythagoras, however we are only asked for the angle to the horizontal `alpha`.

If `sf(theta)` is the angle to the vertical then:

`sf(tantheta=7.517/10.3=0.7298)`

`:.` `sf(theta="tan"^(-1)0.7298=36.12^@)`

So the angle to the horizontal is given by:

`sf(alpha=90^@-36.12^@=53.87^@)`