Choose the correct answer:
1. Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t = 0 s. Ball B is thrown vertically down with an initial velocity u at t = 2 s. After a certain time, both balls meet 100 m above the ground. Find the value of u in ms–1 [use g = 10 ms–2]
Choose the correct answer:
1. Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t = 0 s. Ball B is thrown vertically down with an initial velocity u at t = 2 s. After a certain time, both balls meet 100 m above the ground. Find the value of u in ms–1 [use g = 10 ms–2]
- 10
- 15
- 20
- 30
The Correct Option is D
Solution and Explanation
The correct option is(D): 30 m/s.
Let us assume that they meet at t = t0
\(A:80=\frac{1}{2}gt^2_0....(i)\)
\(B:80=u(t_0-2)+\frac{1}{2}+\frac{!}{2}g(t_0-2)^2...(ii)\)
⇒ 80 = 2u + 5(2)2
⇒ u= 30 m/s
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JEE Main Notification
Concepts Used:
Relative Velocity
The velocity with which one object moves with respect to another object is the relative velocity of an object with respect to another. By relative velocity, we can further understand the time rate of change in the relative position of one object with respect to another.
It is generally used to describe the motion of moving boats through water, airplanes in the wind, etc. According to the person as an observer inside the object, we can compute the velocity very easily.
The velocity of the body A – the velocity of the body B = The relative velocity of A with respect to B
V_{AB} = V_{A} – V_{B}
Where,
The relative velocity of the body A with respect to the body B = V_{AB}
The velocity of the body A = V_{A}
The velocity of body B = V_{B}