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A particle is projected with speed u so as to strike at right angles a plane through the point of projection inclined at 30° with the horizon. The range on this inclined plane is a) 4u^2/7g b) 3u^2/5g c) u^2/2g d) u^2/g

A particle is projected with speed u so as to strike at right angles a plane through the point of projection inclined at 30° with the horizon. The range on this inclined plane is
a) 4u^2/7g  b) 3u^2/5g c) u^2/2g d) u^2/g 

Grade:11

1 Answers

Arun
25750 Points
3 years ago
Let us consider a ball projected at an angle θ with respect to horizontal x-axis with the initial velocity u .
 
The point O is called the point of projection, θ is the angle of projection and OB = horizontal range. The total time taken by the particle from reaching O to B is called the time of flight.
 
Now, 
 
(a). The total time of flight is
 
 Resultant displacement is zero in Vertical direction.
 
Therefore, by using equation of motion
 
s=ut− 21
gt 2
gt=2sinθ
t= g
2sinθ
?(b). The horizontal range is
Horizontal range OA = horizontal component of velocity × total flight time
R=ucosθ× 
g2usinθ
?R= gu2
sin2θ
?(c). The maximum height is
 
It is the highest point of the trajectory point A. When the ball is at point A, the vertical component of the velocity will be zero.
 
By using equation of motion
v 2 =u 
2 −2as
0=u 2
sin 2
θ−2gH
H= 2g
u 2
sin 2θ? 
Hence, this is the required solution

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