Question

A stone is projected from point P on the inclined plane with velocity v= 10 m/s directed perpendicular to the plane.
The time taken by the stone to strike the horizontal ground S is
(Given PO = 10 meter)
(A) 1.5 sec
(A) 1.5 sec
(B) 1.4 sec
(C)2 sec
(D) 2.3 sec
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Answer 1

Answer:

Let the striking angle be x and the reverting angle will be (90-x) from the inclined plane.

Thus by using the formula of range we can find the range of the ball on the inclined plane.

As the range is given in the question as 10 meter hence we can us it to calculate the angle x.

R= u^2(sin(2x)/g.

10=100(sin(2x))/ 10.

10=10(sin2x).

sin2x=1.

sin2x=sin(π/2).

x=π/4.

Hence putting the value of x in the equation of time period(T) we get.

T= 2usinx/g

T=2*10*sin(π/4)/10.

T=$\sqrt{2}$

T=1.4 seconds.