Ball A is thrown vertically up from the ground at time t=0 with an initial velocity of 60m/sec. Ball B is thrown vertically up from ground at t=0 with an initial velocity of 45m/sec. Calculate the maximum distance between the two balls and the time (t) at which this occurs.

Question

Eshan · Answer

From the frame of ball B, ball A moves with up with constant speed of 60m/s-45m/s=15m/s upwards with no relative acceleration. So the distance will constantly increase until the ball B reaches the ground.
Time of flight of ball B= $\dfrac{2u_B}{g}=\dfrac{2\times 45}{10}=9s$
Hence the maximum distance between the two balls in the time= $v_{rel}\times t=15m/s\times 9=135m$

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Ball A is thrown vertically up from the ground at time t=0 with an initial velocity of 60m/sec. Ball B is thrown vertically up from ground at t=0 with an initial velocity of 45m/sec. Calculate the maximum distance between the two balls and the time (t) at which this occurs.

Ball A is thrown vertically up from the ground at time t=0 with an initial velocity of 60m/sec. Ball B is thrown vertically up from ground at t=0 with an initial velocity of 45m/sec. Calculate the maximum distance between the two balls and the time (t) at which this occurs.

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Ball A is thrown vertically up from the ground at time t=0 with an initial velocity of 60m/sec. Ball B is thrown vertically up from ground at t=0 with an initial velocity of 45m/sec. Calculate the maximum distance between the two balls and the time (t) at which this occurs.

Ball A is thrown vertically up from the ground at time t=0 with an initial velocity of 60m/sec. Ball B is thrown vertically up from ground at t=0 with an initial velocity of 45m/sec. Calculate the maximum distance between the two balls and the time (t) at which this occurs.

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