A man throws a stone up into the air at angle θ with the horizontal. a) Draw the path of the projectile and mark the directions of velocity and acceleration at the highest position. (2) b) Derive an expression for the time flight.
Answers
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−21gt2
gt=2sinθ
t=g2sinθ
(b). The horizontal range is
Horizontal range OA = horizontal component of velocity × total flight time
R=ucosθ×g2usinθ
R=gu2sin2θ
(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
v2=u2−2as
0=u2sin2θ−2gH
H=2gu2sin2θ
Hence, this is the required solution
Explanation:
Let us consider a ball projected at an angle e with respect to horizontal x-axis with the initial velocity u.
The point O is called the point of projection, 0 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-21gt2
gt=2sine
t=g2sine
(b). The horizontal range is
Horizontal range OA = horizontal component
of velocity x total flight time
R=ucosexg2usine
R=gu2sin20
(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
v2=u2-2as
O=u2sin20-2gH
H=2gu2sin20
Hence, this is the required solution.