Projectile motion from Moving frame
Learn how projectile motion changes when observed from a moving frame of reference. Explore key concepts, equations, and examples in this beginner-friendly physics guide.
An observer A is stationary on ground. Another observer B is standing in a car which is moving with speed v along horizontal.
1. If B projects the ball vertically upward with speed u then the horizontal and vertical component of velocities of the ball with respect to observer A are respectively ux = v and uy = u.
2. If B projects the ball with speed u at an angle 
with horizontal (away from A), then the horizontal and vertical component of velocities of the ball with respect to observer A are respectively ux = (ucos + v) and uy = usin.
3. If B projects the ball with speed u at an angle with horizontal (towards A), then the horizontal and vertical component of velocities of the ball with respect to observer A are respectively ux = (v – ucos) and uy = usin.
4. If observer B projects the ball with speed u at an angle with horizontal (away from A) from an elevator which is moving upward with speed v, then the horizontal and vertical component of velocities of the ball with respect to observer A are respectively ux = ucos and uy = (usin + v).
5. An observer A is standing on ground. A man B is standing on a trolley which is moving through the horizontal ground with speed v. Man fires a bullet with velocity u with respect to trolley at an angle with horizontal. If the friction between trolley and ground is negligible then calculate the range of the bullet with respect to observer A (considered when the bullet was fired, A and B are at same position).
The horizontal component of initial velocity of bullet with respect to observer A is ux = ucos + v ——– (i)
The vertical component of initial velocity of bullet with respect to observer A is uy = usin ——- (ii)
At any instant of time t, the horizontal distance travelled by the bullet with respect to A is x = uxt = (ucos + v)t ——- (iii)
At time t, the vertical displacement of bullet with respect to A is y = uyt – 
= utsin – ——- (iv)
When the bullet touches the ground then vertical displacement of it is zero.
Therefore, from equation (iv) we get, utsin – = 0 or, t = 
.
The range of the bullet with respect to observer A is R = 
. [From equation (iii)]
6. A man is standing on a car which is moving with speed 2u on a frictionless horizontal surface. The man throws a ball with a speed u at any angle 
with respect to the car. At the moment when the man throws the ball, driver of the car accelerates it with a in the initial direction of its motion and finally the man catches the ball. Calculate the angle of projection of the ball and the time after when the man catches the ball.
The horizontal component of initial velocity of the ball is ux = ucos + 2u ——– (i)
The vertical component of initial velocity of the ball is uy = usin ——- (ii)
Let us consider the man throws the ball from point A and catches it at point B.
At any instant of time t, the horizontal and vertical displacement of the ball with respect to ground are respectively x = uxt = (ucos + 2u)t ——- (iii) and y = uyt – = utsin – ——- (iv)
At point B the vertical displacement of the ball is zero. So, utsin – = 0 or, t = 
.
The distance travelled by the car AB = 2ut + 
The range of the ball is R = uxt = (ucos + 2u)t
Therefore, (ucos + 2u)t = 2ut +
Or, 2ucos = at
Or, cos = 
= 
[Putting the value of t]
Or, tan = 
∴ = 
.
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