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Question

Two balls are thrown from an inclined plane at angle of projection α with the plane, one up the incline and other down the incline as shown in figure (t stands for total time of flight):


A

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B

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C

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D

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Solution

The correct options are
A


B


C


Range over the incline is the distance travelled along the plane in the given time period.

Sx=uxt+12axt2 Sx=uxt+12axt2

R1=vocos αT12gsin θT2 R2=vocos αT12gsin θ T2

R2R1=gsin θ T2

So option C is correct

vt1&vt2 are the velocities of the particles at their maximum heights. Let the particles reach their maximum heights at time t1 and t2 respectively. Hence

vy=uy+ayt

0=(v0sin α)(gcos θ)t1t1=vosin αgcos θ

Similarly t2=vosin αgcos θ Hence t2=t1

Hence vt1=vocos α(gsin θ)t1

vt2=(vocos α)(gsin θ)t2vt1vt2

Range over the incline is the distance travelled along the plane in the given time period.

Sx=uxt+12axt2 Sx=uxt+12axt2

R1=v0cos αT12gsin θT2 R2=vocos α T12gsin θ T2

R2R1=gsin θ T2

So option C is correct

vt1&vt2 are the velocities of the particles at their maximum heights. Let the particles reach their maximum heights at time t1 and t2 respectively. Hence

vy=uy+ayt

0=(vosin α)(gcos θ)t1t1=vosin αgcos θ

Similarly t2=vosin αgcos θ Hence t2=t1

Hence vt1=vocos α(gsin θ)t1

vt2=(vocos α)(gsin θ)t2vt1vt2


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