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PhysicsKinematics1-DMedium2 minPYQ_2022
PhysicsMediumnumerical

From the top of a tower, a ball is thrown vertically upward which reaches the ground in6 s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in1.5 s. A third ball released, from the rest from the same location, will reach the ground in _____ s.

Answer:
3.00
Solution:

When it is thrown upward, -h=v×6-12g62

When it is thrown downward, -h=-v×1.5-12g1.52

Therefore, 5h=536+9 h=45 m

When the ball is dropped, h=12gt245=12×10×t2t=3 s

Alternate method, t=t1t2=6×1.5=3 s

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:1-D
2mℹ️ Source: PYQ_2022

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