41

A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with an acceleration of 1 m/s² and the projection velocity in the vertical direction is 9.8 m/s. How far behind the boy will the ball fall on the car?

aNSWE"
we shall use the following kinematic equation to describe the vertical motion of the ball to to its maximum height (half its journey in flight)
v = u + at

here
v = o,u = 9.8 m/s,a = -9. m/s2

thus,
0 = 9.8 - 9.8t
so, the time taken by the ball to reach maximum height will be
t = 9.8 / 9.8 = 1 secs.
or total time of flight, T = 2t = 2 secs

now, the horizontal distance travelled by car in 2 seconds will be
s = uT + (1/2)aT2

here,
u = 0,T = 2 secs.,
a = -1 m/s2

thus,
s = 0 + (1/2).-1.22

so, the distance between ball and car will be

s = -2m

aLTERNATE:-

Sol. Let the velocity of car be u when the ball is thrown. Initial velocity of car is = Horizontal velocity of ball. 
Distance travelled by ball B Sb = ut (in horizontal direction) 
And by car Sc = ut + 1/2 at2 where t → time of flight of ball in air. 
∴ Car has travelled extra distance Sc – Sb = 1/2 at2. 
Ball can be considered as a projectile having θ = 90°. 
∴ t = (2u sinθ)/g = (2 x 9.8)/9.8 = 2 sec. 
∴ Sc – Sb = 1/2 at2 = 2 m 
∴ The ball will drop 2m behind the boy. 

alternate:-

41. A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with an acceleration of 1 m/s² and the projection velocity in the vertical direction is 9.8 m/s. How far behind the boy will the ball fall on the car ?    



Answer: First of all we should know the time duration for which the ball is in the air. Let this time be t, vertical displacement in this time is zero. and given that u=9.8 m/s. From the equation h=ut-½gt², We get,

0=9.8t-½9.8*t²,

→ ½t²-t=0, 

→ t(t-2)=0, → t=0 or 2 s. t=0 corresponds to the time when it is being projected. So the time duration for which it remains in air = 2 s.

      Since the ball is projected straight upwards its horizontal velocity with respect to the frame of railroad car is zero at the instant of projection. But the frame is itself moving with an acceleration of 1 m/s² in horizontal direction so with respect to this accelerating frame the ball will have equal but backwards acceleration during the flight. So to calculate the horizontal displacement with respect to railroad car we have following data,

u=0, a=-1 m/s², t=2 s, So displacement x=ut+½at²  

=0-½*2² =-2 m. Negative sign shows backwards displacement. 

So the ball will fall 2 m behind the boy on the car.