Example of two dimensional motions | Learn Physics

Example of two dimensional motions

Question: A very broad elevator is going up vertically with a constant acceleration of 2 ms^-2 . At the instant when its velocity is  4ms^-1 a ball is projected from the floor of the lift with a speed of 4ms^-1 relative to the floor at an elevation of 30ᵒ .

If g = 10 ms^-1  , find (a) the time taken by the ball to return to the floor. (b) The range of the ball over the floor of the lift.

Solution:

Here , u = 4 m/s, θ = 30ᵒ.

Acceleration of the ball relative to the lift

=10+2 = 12 m/s²

Acting in the negative y-direction or vertically downwards.  It means’

Here   g’=12 m/s²

Time of flight, T = 2u sin θ /g’=2 × 4 × sin30ᵒ/12=1/3 s

Horizontal range, R = u² sin 2θ /g’= (4)² sin 2 × 30ᵒ/12= 1.15 m

 

Question: A cannon ball has the range R, for two angles of projection. If the maximum height attained are H and H’ respectively and t and t’ be the respective times of flight in two possible paths, prove that R = 4 × (HH’)^1/2 = ½ gtt’.

 Solution:

A cannon ball when projected with a velocity u can have the same range R if its angle of projection is θ or (90ᵒ - θ) from the horizontal. So

R =  u² sin 2θ /g = u² sin 2(90ᵒ - θ) /g           

For the first case,  H = u² sin ²θ /2g                 ……………..(1)

T = 2u sin² θ/g                                                     ……………(2)

For the second motion,    H’ = = u² sin²(90ᵒ - θ) /2g = u² cos ²θ /2g    ………(3)

t ‘ = 2u sin(90ᵒ - θ) /g = 2u cos θ/g    ……………………….(4)

From (1) and (3)  , HH’ = u⁴ sin ²θ cos²θ/4g² =( 1/16)R²

R² = 16 HH’ or R = 4(HH’)^1/2

From (2) and (4),  tt’  = 4u² sin θ cosθ/g² = 2/g(u² sin 2θ) /g = 2R/g

Or R = ½ gt t’