Suppose a cannonball is fired into the air with an initial velocity of 35.0 m/s

Question

Suppose a cannonball is fired into the air with an initial velocity of 35.0 m/s
at an angle 32? above the horizontal. Ignore the effects of air friction.
(A) How much time will it take the ball to reach the highest point in its trajectory?
(B) How far from the cannon, horizontally, is this point?
(C) At the highest point in its trajectory, the cannonball explodes, breaking into two
parts of equal mass. One of the parts drops stright down. What is the velocity of the
other part, immediately after the explosion?
(D) How far from the cannon will the second

Explanation / Answer

sol t=35sin(32)/9.8 =1.9 sec x=35cos32 *1.9 =56.4 m velocity=59.36 m/sec distance from the top point=59.36*1.9=112.784m distance from canon=112.784+56.4 =169.184 m

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