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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,
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
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