A ball is dropped from the tower 30m high. at the same time second ball is thrown upward from the ground with an initial velocity of 15m/s .when and where do they cross and with what relative velocity?
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when they will meet. their time will be same.
eq.
1/2at^2=30-(ut -1/2at^2)
1/2*10*t^2= 30-15t+1/2*10t^2
5t^2=30-15t+5t^2
15t=30
t=2s
they will meet 2s after their motion
10 m high from the ground
falling ball v=20m/s
thrown ball v=5m/s
relative velocity = 15m/s at that point
due to their same direction of motion.
eq.
1/2at^2=30-(ut -1/2at^2)
1/2*10*t^2= 30-15t+1/2*10t^2
5t^2=30-15t+5t^2
15t=30
t=2s
they will meet 2s after their motion
10 m high from the ground
falling ball v=20m/s
thrown ball v=5m/s
relative velocity = 15m/s at that point
due to their same direction of motion.
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