A projectile is projected upwards from level ground, with an initial velocity u and angle of projection 60' with horizontal. It is found that at a particular position P on its path the radial distance from the point of projection O is twice the vertical height of P. The distance of OP is:
Answers
ANSWER
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−
2
1
gt
2
gt=2sinθ
t=
g
2sinθ
(b). The horizontal range is
Horizontal range OA = horizontal component of velocity × total flight time
R=ucosθ×
g
2usinθ
R=
g
u
2
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
2
θ
Hence, this is the required solution
Answer:
check the attached file
