A very broad elevator is going up vertically with a constant acceleration of 2m/s2. At the instant when its velocity is 4m/s a ball is projeted from the floor of the lift with a speed of 4m/s relative to the floor at an elevation of 30∘. The time taken by the ball to return to the floor is (g=10m/s2)
Answers
Answer:
Explanation:
=> when the velocity of elevator is 4m/s, a ball is projeted from the floor of the lift with a speed of 4m/s relative to the floor at an elevation of 30°.
Thus, elements of velocity of ball as respect to lift :
u_x = 4 * cos30°
= 4 * √3 / 2
= 2√3 m/s
u_y = 4 sin30°
= 4 * 1 /2
= 2 m/s
=> According to the question, a very broad elevator is going up vertically with a constant acceleration of 2m/s2. Thus, ball's acceleration as compared to lift in negative y direction (downwards) is:
a= 10 + 2
=12 m / s²
=> Hence, the time taken by the ball to return to the floor:
T=2*u_y / a
=2 * 2 / 12
= 4 / 12
=1 / 3
= 0.33 s
=> Therefore, the time taken by the ball to return to the floor is 0.33 s.
Given :-
Acceleration of ball = a(b) = -10 ms-²
Acceleration of elevator = a(e) = -2 ms-²
Angle of Projection = Ø = 30°
Relative Acceleration of ball w.r.t elevator.
a(be) = -10 + (-2)
a(be) = -12 ms-².
T = 2u SinØ/g
T = (2 × 4 × 1/2)/12
T = 4/12
T = 1/3 s
Hence,
Time taken = T = 1/3 s.