A particle is released from the top of a tower of height 150 m and another ball is thrown vertically upward with velocity 30 m/s from the bottom of the tower. Assume both are in a same vertical line, the time after which they collide each other
Question
A particle is released from the top of a tower of height 150 m and another ball is thrown vertically upward with velocity 30 m/s from the bottom of the tower. Assume both are in a same vertical line, the time after which they collide each other
Solution
Let's denote:
- The time it takes for the two balls to collide as t.
- The height of the tower as H = 150 m.
- The initial velocity of the ball thrown upwards as u = 30 m/s.
- The acceleration due to gravity as g = 9.8 m/s² (downwards).
The ball thrown upwards will move against gravity, so its height at any time t can be given by the equation of motion: h1 = ut - 0.5gt².
The ball dropped from the tower will only be under the influence of gravity, so its height at any time t can be given by the equation: h2 = H - 0.5gt².
The two balls will collide when h1 = h2. So, we can set the two equations equal to each other and solve for t:
ut - 0.5gt² = H - 0.5gt²
Solving for t gives:
t = H / u
Substituting the given values:
t = 150 m / 30 m/s = 5 s
So, the two balls will collide 5 seconds after they are released/thrown.
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