REST AND MOTION 37

Q 37. A person standing on the top of a cliff 171 ft high has to throw a packet to his friend standing on the ground 228 ft horizontally away. 

If he throws the packet directly aiming at the friend with a speed of 15.0 ft, how short will the packet fall ?

The person standing at A throws the packet along AC with a velocity u = 15 ft/s. Let the angle of velocity with horizontal be θ. 

Now tan θ = 171/228 = 3/4, This gives sin θ = 3/5  and  cos θ =4/5.

 Vertical component of the velocity =u sin θ = 15*3/5 = 9 ft/s. This is the initial velocity of packet for vertical accelerated motion. If the time to reach the ground be t. Using the equation h=ut+½gt² , we get, 

171 = 9t+½*32*t² → 16t²+9t-171=0 →t={-9±√(9²+4*16*171)}/32

taking only + ve sign for t, we get, t = (-9+105)/32 = 3 s.

 In this time the packet will travel with horizontal component of the velocity uniformly.

Horizontal component of velocity = u cos θ = 15 *4/5 =12 ft/s. 

So the horizontal distance BD = 12 ft/s x 3 s = 36 ft. So the packet falls a distance CD ft short of C. CD = BC - BD = 228 -36 = 192 ft