.A boy throws a cricket ball with a velocity u at angle with the horizontal.
(a)Name the path followed by the ball.
(b)At the higest point, what are the vertical and horizontal component of velocity?
(c)Obtain an expression for horizontal range and write condition for maximum horizontal range. (d)What are the direction of acceleration of the body at different points?
(e)If 0,=30°, then what is the value of 0₂? U₂ ← Horizontal range
(f)Derive an expression for the maximum height reached by the ball.
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