Show that velocity with which a projectile hits the ground is same with which it was initially projected and angle made by the trajectory with the horizontal is same as the angle of the projection
Answers
Consider a projectile with initial velocity u and projected at an angle θ with the horizontal x-axis.
The velocity and acceleration has two components. Horizontal (x – axis) and vertical (y axis).
Initial velocity = u
Horizontal component of initial velocity = ux = ucos(θ)
Vertical component of initial velocity = uy = usin(θ)
Horizontal component of acceleration = ax = 0
Vertical component of acceleration = ay = -g.
Since horizontal acceleration is zero, the horizontal velocity remains same until end.
Vertical velocity vy = uy + ay*t
vy = usin(θ) - gt. -------------------------------------E1
To calculate t, time period of projectile, consider the vertical component.
The vertical displacement = 0
Substituting values in S = ut + at²/2, we get
0 = usin(θ)t – gt²/2
t = 2usin(θ)/g
Substituting t value in E1, we get
Vy = usin(θ) – g*2usin(θ)/g = -usin(θ)
That means , verticle velocity is in opposite direction.
Initial velocity = usin(θ) + ucos(θ)
Final velocity = -usin(θ) + ucos(θ)
So the scalar component of both velocities are same. Only difference is vertical direction is opposite.
Initial angle of the projectile is θ with the ground.
The θ is result of horizontal and vertical components. From above calculation we proved that only vertical component of velocity will become opposite direction.
So final angle of the projectile also same with ground except the direction is different.
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