Question

A particle is thrown horizontally with a speed of 10 m/s, from the top of a cliff
15 m high above the ground. An inclined plane runs at an angle 45° with the
horizontal as shown in the figure. At what height above the
ground does the particle hit the plane? (Take g = 10 m/sec)​

Answer 1

Answer:

10 m

Explanation:

A particle is thrown horizontally with a speed of 10 m/s, from the top of a cliff

15 m high above the ground. An inclined plane runs at an angle 45°

Let say after t sec particle hit inclined plane

Then Horizontal distance covered = 10t  m

10t horizontal distance => 10t height above the ground

Vertical distance covered = 0 + (1/2)gt² = 5t²

Vertical distance covered + height above the ground = 15

=> 5t² + 10t = 15

=> t² + 2t = 3

=> t² + 2t - 3 = 0

=> t² + 3t - t - 3 = 0

=> t(t + 3) -1(t + 3) = 0

=> (t - 1)(t + 3) = 0

=> t = 1 sec

10t = 10 m

10 m is the height above the  ground when the particle hit the plane