Question

A boy is sliding down an inclined plane of
inclination theta. He throws a ball in such a
direction and with such a velocity that he
can catch it after a time T seconds. Then the
velocity of the ball w.r.t. himself was
1) Tgcos(theta)
2) Tgsin(theta)
3) Tg sin(theta)/2
4) Tgcos (theta)/2
ans:4, explanation please :))
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Answer 1

Answer:

From Newton's second law for the bar in projection from, Fx=mwx along x direction we get

mgsinα−kmgcosα=mw

or, vdvdx=gsinα−axgcosα, (as k=ax),

or, vdv=(gsinα−axgcosα)dx

or, ∫v0vdv=g∫x0(sinα−xcosα)dx

So, v22=g(sinαx−x22acosα) (1)

From (1) v=0 at either

x=0, or x=2atanα

As the motion of the bar is unidirectional it stops after going through a distance of 2atanα.

From (1), for vmax,

ddx(sinαx−x22acosα)=0, which yields x=1atanα

Hence, the maximum velocity will be at the distance, x=tanα/a

Putting this value of x in (1) the maximum velocity,

vmax=gsinαtanαa−−−−−−−−−−√

TRY IT OUT