the ball is dropped from top of the tower, after 2 seconds another ball is thrown vertically downward with a speed 40m/sec after how much time and at what distance the ball meets
Answers
Answer:
Given information,
Total height, h = 50m
Acceleration, g = 10 m/s^2
For ball 1-
Time, t = t s
Initial velocity, u = 0 m/s
=> s = ut + (1/2)at^2
=> s = 0t + (1/2)10t^2
=> s = 5t^2
For ball 2-
Time, t = (t-2) s
Initial velocity = 40 m/s
Acceleration, g = -10 m/s^2 (-ve sign because in opposite direction)
Initial velocity, u = 40 m/s
=> s2 = ut + (1/2)at^2
=> s2 = 40(t+2) + (1/2)(-10)(t-2)^2
=> s2 = 40t + 80 + (-5)(t^2 - 4t + 4)
=> s2 = 40t + 80 - 5t^2 + 20t - 20
=> s2 = -5t^2 + 60t +60
Ball 2 covered s2 distance from bottom.
=> s1 + s2 = 50
=> 5t^2 -(- 5t^2 + 60t + 60) = 50 (because s1 and s2 are in opposite direction)
=> 5t^2 + 5t^2 - 60t - 60 = 50
=> 10t^2 -60t -110 = 0
=> t^2 - 6t - 11 =0
Answer:
the ball is dropped from top of the tower, after 2 seconds another ball is thrown vertically downward with a speed 40m/sec after how much time and at what distance the ball meets
Explanation:
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