A ball is projected vertically upward and bounces elastically off a roof inclined at 45° at a height 10 m above the projection point; it then lands on a table 5 m high at a horizontal distance of 10 m

Question

A ball is projected vertically upward and bounces elastically off a roof inclined at 45° at a height 10 m above the projection point; it then lands on a table 5 m high at a horizontal distance of 10 m from the start. The projection speed is: (Take g = 10 m / s 2 )

TestHub · Accepted Answer

Fall from 10 m to 5 m : t = 10 2 × 5 ​ ​ = 1 s, so the post-bounce horizontal speed is 1 10 ​ = 10 m / s ; v 0 ​ = 1 0 2 + 2 × 10 × 10 ​ = 300 ​ = 10 3 ​ m / s .

Physics - Kinematics Question with Solution | TestHub

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