Question

Q) Ball A is thrown straight up with an initial velocity V0 from a cliff of height h above ground, where h is not known. at the same time ball B is dropped from the same cliff. the time of flight of ball B is 1/4 that of ball A. take the upward direction is positive and

A) draw a single V vs. t plot for both balls.

B) draw a single position vs. time plot for both the balls.

c) find the height of the cliff in term of V0 and g.

d) find time of flight for each ball in term of V0 and g.

Answer 1

A) The equation of motion for the ball A is as follows: The graph of velocity versus time graph for the ball A is as follows: Fig : v versus t graph for ball A. For the ball B, -gt , because, the ball B is at rest, so, its initial velocity s equal to zero The graph of velocity versus time graph for the Ball B is as follows: Fig 2: v versus t graph for ball B

B) The equation of motion for the position of ball A is as follows: The graph of position versus time for the ball A is as follows: Fig 3: v versus t graph for ball A. The equation of motion for the position of ball B is as follows: gt The graph of position versus time for the ball B is as follows: Fig 4: v versus t graph for ball B.

C) Calculate the final displacement as follows 294 gt 200= gt-16

Solve for t 2vo 16 g 15 32v 15g Now, calculate the height as follows (32%)2 1 2 60g 512 ( (16)(225)(:g 32 225人g The time taken by the ball A is, 32v 15g And, for the ball B is 15g