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.

## Comments

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TopNewestFIITJEE? Cool. – Nishanth Hegde · 4 years, 3 months ago

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– Rohan Rao · 4 years, 3 months ago

Yep...I recently joined FIITJEE for IIT coaching. Have you joined any coaching classes for IITJEE/Medical Entrances?Log in to reply

– Nishanth Hegde · 4 years, 3 months ago

BASE IIT-JEE achiever programmeLog in to reply

– Rohan Rao · 4 years, 3 months ago

Ah...I see...I was at BASE for class 9 and 10. The teachers were very good!Log in to reply

Wait. Shouldn't time period be ucos(theta)/gsin(theta)? How did you get your answer? – Aditya Karekatte · 4 years, 3 months ago

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– Rohan Rao · 4 years, 3 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...:DLog in to reply

The time period is t=ucos(α)/gsin(β),you're changing v for u in the equation that you did. – Rafael Saboya · 4 years, 3 months ago

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– Rohan Rao · 4 years, 3 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.Log in to reply

– Rafael Saboya · 4 years, 3 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.Log in to reply