Question

A stone is projected horizontally from a point P., so that it hits the inclined plane perpendicularly. If the stone is at a vertical height h from the foot of an inclined plane which h is inclined at an angle theta with the horizontal. Find v0(velocity of projection of the stone) ​

Answer 1

Answer:

Explanation:

Taking components along all directions such as x and y.

Let the angle of inclination of the inclined plane be m.

For velocity along x direction.

u(x)=vo*cos(m)

And, acceleration along x axis.

a(x)= - gsin(m)

For velocity along y dirrection.

u(y)= vo*sin(m)

For acceleration along y dirrection.

a(y)= gsin(m)

Using newtons laws of motion,

u(x) + a(x)t= 0

t= {vo*cos(m)}/gsin(m)

The distance s(y)= hcos(m)

u(y)t + 1/2*a(y)*t^2 = hcos(m)

{vo*sin(m)}* {vo*cos(m)}/gsin(m) + 1/2*gcos(m)* [{vo*cos(m)}/gsin(m)]^2

=hcos(m)

On solving we get,

vo = ✓(2gh)/(1+cot^2(m))

Answer 2

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