Two balls are thrown simultaneously, one vertically upward from the ground and the other downward from a height of 40 m. If both are thrown with same velocity, at what height from the ground will they collide?(g=9.8 m/s2).

1
by amrithabhi

2014-09-01T23:18:57+05:30

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Look at the diagram for clarity. Ball2 is thrown upwards with a speed = U m/sec and Ball1 is thrown downwards with a speed of U m/sec. Let Ball1 and Ball2 meet at a height D from ground at t = T sec after the balls are thrown.

Distance traveled by ball1 from height of 40m:    40 - D = U T + 1/2 9.8 T²
40 - D = U T + 4.9 T²    --  Equation 1

Distance traveled by ball 2 from ground upwards:      D = U T - 1/2 9.8 T²
D = U T - 4.9 T²    --  equation 2

Adding two equations, we get:  40 = 2 U T    => U T = 20  --  equation 3
So T = 20/U sec.

4.9 T² = 20 - D    from equation 2 and 3.
4.9 (20/U)² = 20 - D

D = 20 - 1960/U²

The distance D depends on the value of U. If U is given then we can find D.

If U = 0 m/s ,       D = 0 m/s
if U = 10.15 m/s   D = 1 m
if U = 44.27 m/s , D = 19 m
if U = 14m/s         D = 10 m
if U = infinity,        D = 20 m