# Projectile on an Incline

A particle is projected with velocity u from the bottom of an inclined plane whose inclination with the horizontal is $$\beta$$. If afterwards the projectile strikes the inclined plane perpendicular to the surface, find the height of the point struck (distance from the ground). The angle made by the velocity with the incline is $$\alpha$$.

I got an answer while solving this question that differs from the one in my textbook. I got the correct value for the time period though...could someone please try this question? Please help me find the height and also tell me the expression you get for the time taken. Thank you! :D

Go to the website http://physics.stackexchange.com/questions/62647/projectile-along-incline for the working I did.

Note by Rohan Rao
5 years, 6 months ago

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- 5 years, 6 months ago

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- 5 years, 6 months ago

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- 5 years, 6 months ago

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- 5 years, 6 months ago

Wait. Shouldn't time period be ucos(theta)/gsin(theta)? How did you get your answer?

- 5 years, 6 months ago

Aditya, check the image uploaded on Physics.SE. The link is above, in the discussion. My method differs from the conventional method of vector components along the incline, as I have mentioned on Physics.SE. However, the answer obtained is correct and the same in both cases. There is no doubt about the time period. Btw, check your expression...what do you mean by theta? I have not supplied such an angle...:D

- 5 years, 6 months ago

The time period is t=ucos(α)/gsin(β),you're changing v for u in the equation that you did.

- 5 years, 6 months ago

This is also a correct expression for the time period. In fact, both expressions have been equated in my textbook. But now what I need to know how to find is the height.

- 5 years, 6 months ago

i see,i took a rapid view at the question... you tried to use the relation between the angles? your answer may be right,just needing some trigonometry.

- 5 years, 6 months ago