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.

1 Answer
Michael
Feb 27, 2017

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

#sf(53.9^@)#

Explanation:

MFDocs

#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^@)#

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