A slingshot is used to launch a stone horizontally from the top of a 25.0 meter cliff. The stone lands 33.0 meters away. A. At what speed was the stone launched? B. What is the speed and angle of impact?

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Answer 1 · 2017-12-12T15:15:53

26.5 m/s at an angle of #sf(56.7^@)# to the horizontal

Considering the vertical component of the motion we get:

#sf(h=1/2"g"t^2)#

#:.##sf(t=sqrt((2h)/(g))#

#sf(t=sqrt((2xx25.0)/(9.81))=2.26color(white)(x)s)#

Now we can find #sf(v_x)# which is constant:

#sf(v_x=33.0/2.26=14.6color(white)(x)"m/s")#

Now we consider the vertical component of the motion using:

#sf(v_(y)=u_(y)+at)#

This becomes:

#sf(v_(y)="g"t)#

#sf(v_(y)=9.81xx2.26=22.17color(white)(x)"m/s")#

To find the resultant #sf(R)# of #sf(v_x)# and #sf(v_y)# we use Pythagoras:

#sf(R=sqrt(22.17^2+14.6^2)#

#sf(R=26.54color(white)(x)"m/s")#

To find the angle #sf(theta)# to the horizontal which the stone strikes the ground we get:

#sf(sintheta=v_y/R=22.17/26.54=0.835)#

#sf(theta=56.7^@)#