 # A body thrown along a frictionless inclined plane of angle of inclination 30 degrees covers a distance 40m along the plane. If the body is projected with the same speed at an angle of 30 degree with the ground, it will have a range of(take g=10m/s) 8 years ago
First we need to calculate the velocity of the body on the inclined plane. For that we use the formula v^2=u^2+2as. We cannot use the value of acc 10 m/s^2 because acc is a vector quantity we cannot add it normally.for that we break the component of acc along the inclined plane.we have a=g cos theta=10root3/2. Now,v^=2*10*root3/2*40=400root3. Now,we need to calculate the horizontal range for that we use the formula R=u^2sin2theta/g=(400root3*sin120/10=400root3*root3/2)/10=60m This is the required range. Hope this helps
8 years ago
Sorry I have taken a wrong component of acc it should actually be g sin theta while dealing with the velocity on incline.rest things are correct you can take acc as g sin theta and solve the question the same way as I have done above.
8 years ago
First we use the formula (V^2)-(U^2)=2as here acceleration due to gravity acts on a body.and the body decelerates. therefore a=-g. and after traveling 40 m the body stops,therefore final velocity v=0 -(U^2)=-2g(40) formula is (ucos30)^2=80g therefore u^2=320g/3--->1 R=u^2sin2(30)/g =320*g*sin60/3g =160/v3=92.3
8 years ago
First we use the formula (V^2)-(U^2)=2as here acceleration due to gravity acts on a body.and the body decelerates. therefore a=-gcos30. and after traveling 40 m the body stops,therefore final velocity v=0 -(U^2)=-2gcos30(40) formula is (ucos30)^2=80gcos30 therefore u^2=160g/v3--->1 R=u^2sin2(30)/g =160*g*sin60/v3g =(160*v3)/(2v3)=80m
8 years ago
I think some information is missing in the question,it is not given wether the body travels up the incline or down the incline depending on it we need to take the value of acc -g sin theta or +g sin theta and do further calculations of the velocity
5 months ago

First by using the first part of the question, we have to find the initial velocity which the body was thrown (u).

Here we can use Energy conservation law to solve this,

If we take the Potential energy is 0 when the body was thrown from the bottom of the inclined plane we can write,

Kinetic Energy when the body was thrown from the bottom of the inclined plane = Potential Energy when the body stops.

1/2 x m x u^2 = m x g x 40Sin(30)

u = 20 m/s

Then we use this initial velocity to find the answer by using Linear Motion Equations,

We can first apply s = ut + 1/2at^2 for vertically upward direction,

0 = 20Sin(30) x t + 1/2 x (-10) x t^2

0 = 10t - 5t^2

By solving this quadratic equation we can get,

5t(2-t) = 0

t(2-t) = 0

t = 0 / t = 2

This gives the two answers for t where s = 0,

• Before throwing the body s = 0 , t = 0
• After the body landed on the surface s = 0 , t = 2

So we should take t = 2

Then we can apply s = ut + 1/2at^2 horizontally ---> direction,

s = 20Cos(30) x 2 + 1/2 x 0 x 2^2 , there is no acceleration for that direction

s = 20√3/2

s   = 20√3