A ball is thrown upward with velocity 20m/s from the top of the tower having a height of 50 m. What time elapsed when it again passes edge of tower? (G=10 m/s^2)
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Answers
Answered by
6
Answer:
4s
Explanation:
u = 20 m/s
v = 0 (at the highest point)
a=-10m/s^2
v = u + at => t = v - u/a = 2s
t(up) =2s
=> the ball takes 2 s to reach the highest point
Wkt time taken to move up from a point is equal to time taken to return to same point
t(down) =2s
Total time taken by the ball to reach the same point is t(up) +t(down)
=4 s
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Answered by
4
Answer:
4 sec
Explanation:
Time of flight:
T=2u/g
u=20m/sec
Therefore,
T= 2×20/10= 4sec
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