A rock is shot from a slingshot standing 2.50 m above the ground. It hits a man

Question

A rock is shot from a slingshot standing 2.50 m above the ground. It hits a man standing on a 3.60 m high pillar, 14.0 m away. The initial velocity of the rock is 15.9 m/s. Assume that . Calculate the angle that the rock is launced at, you can use Newtonian Iteration. At what velocity (magnitude and direction) does the rock hit the man ( magnitude and direction) ? If you got two angles in section-a) then you should use the larger angle here I threw together a picture to explain the problem more clearly:

Explanation / Answer

x = xo + vxo t + 1/2 ax t2

y = yo + vyo t + 1/2 ay t2

from above

x0=0

y0=2.5

Vx0=15.9cos(theta)

Vy0=15.9sin(theta)

substitute

x=14

y=3.6

14=15.9cos(theta)t

ax=0

3.6=2.5+15.9sin(theta)t+5t^2

1.1=15.9sin(theta)t+5t^2

sin(theta)=(1.1-5t^2)/15.9t

cos(theta)=14/15.9t

solving these equations

(1.1-5t^2)/15.9t=(1-(14/15.9t)^2)^1/2

(1.1-5t^2)^2=(15.9t)^2-14^2)

t=3.34s

cos(theta)=14/53.1

theta =74 degrees

velocity

Vx is constant

Vy=15.9*0.96

=15.26

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