Exams › JEE Advanced › Physics

A person standing on the roof of a 40 m high tower throws a ball vertically upward with a speed of 10 m/s. Two seconds later, the person throws a second ball in the vertical direction. Both balls hit the ground at the same time. Which of the following statements is/are correct?

1. The first ball hits the ground after 4 seconds.
2. The second ball was projected vertically downwards with speed 10 m/s.
3. The first ball was projected vertically upwards with speed 10 m/s.
4. The second ball was projected vertically upwards with speed 10 m/s.

Correct answer: The second ball was projected vertically downwards with speed 10 m/s.

Solution

Ball 1 hits the ground after 4 s (solving -40 = 10t - 5t²). Ball 2 is thrown 2 s later and must cover -40 m in 2 s: -40 = u*2 - 5*4 gives u = -10 m/s, i.e. downward at 10 m/s.

Related JEE Advanced Physics questions

• A person riding in a hot-air balloon experiences an upward acceleration of 4.9 m/s². Exactly 2 seconds after the balloon is released from ground level, the person drops a small object. Taking g = 9.8 m/s², find the maximum height above the ground that the dropped object reaches.
• An object is released from rest at some height above the ground. In addition to gravity, air exerts a constant horizontal acceleration on the object throughout its fall. The path of the object as seen from the ground frame will be a
• A swimmer can cross a still river of width d to a point exactly opposite the starting bank in time t0. When the river has a current, the swimmer still manages to reach the directly opposite point (same path), but now takes time t1 > t0. What is the speed of the river current?
• A car can produce a maximum acceleration of 5 m/s² and its brakes can produce a maximum retardation of 10 m/s². Assuming the car starts from rest and must come to rest at a destination 1500 m away, find the minimum time (in seconds) to travel this distance.
• From the top of a 100 m tower, a particle is thrown vertically upward with speed 50 m/s at t = 0. Two seconds later, a second particle is projected vertically upward from the ground with the same speed 50 m/s. At what time after t = 0 are the two particles at the same height?
• The velocity of a particle traveling in a straight line is given by v(t) = 5 - 6*e^(-t/2) m/s, where time t is in seconds and t >= 0. At t = 0, the particle is at position x = 7 m. The position as a function of time is x(t) = k*t + l*e^(-t/2) + m. Find the numerical value of (k + m) / l.

⚔️ Practice JEE Advanced Physics free + battle 1v1 →