A person is riding on a flatcar traveling at a constant speed of 200 km/s (Fig.). He wishes to throw a ball through a stationary hoop 90 km above the height of his hands in such a manner that the ball will move horizontally as it passes through the hoop. He throws the ball with a speed of 250 m/s with respect to himself.
(a) How many time required after he releases the ball will it pass through the hoop?
(b) What must the vertical component of the initial velocity of the ball?
(c) At what horizontal distance in front of the hoop must he release the ball?
(d) When the ball leaves the man's hands, what is the direction of its velocity relative to the
frame of reference of the flatcar? Relative to the frame of reference of an observer standing
on the ground?
Answers
Answer: hope it is helpful to youplease mark me as brainlist...
Explanation:
12.5m
15.5m
17.5m
20m
Answer :
C
Solution :
Two important aspects to be noticed in this problem are: (1) Velocity of projection of ball is relative to man in motion
(2) Ball clears the hoop when it is at the topmost point
V→ball,man=V→ball−V→man,V→ball=V→ball,man+V→man
Now we apply the above relation to x-as well as y-component of velocity.If ball is projected with velocity v0 and angle θ, then x-component of
V→ball=(v0cosθ+10)m/s y-component of V→ball=(v0sinθ)m/s.Since vertical component of ball's velocity is unaffected by horizontal motion of car, we can use formula for time of flight.
i.e., (12.5sinθ)22g=5m,orsin2θ=5×(2×10)12.5×12.5
or sinθ=45andcosθ=35
and v0sinθ=(12.5)×(45)=10m/s
Time taken to reach maximum height =2v0sinθg
=2×1010=2 seconds
Horizontal distance of loop from point of projection
=(12.5cosθ+10)×1=17.5m