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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):

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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 αT−12gsin θT2 −R2=vocos αT−12gsin θ T2

R2−R1=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)t1⇒t1=vosin αgcos θ

Similarly t2=vosin αgcos θ Hence t2=t1

Hence vt1=vocos α−(gsin θ)t1

vt2=(−vocos α)−(gsin θ)t2⇒vt1≠vt2

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

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

R1=v0cos αT−12gsin θT2 −R2=vocos α T−12gsin θ T2

R2−R1=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)t1⇒t1=vosin αgcos θ

Similarly t2=vosin αgcos θ Hence t2=t1

Hence vt1=vocos α−(gsin θ)t1

vt2=(−vocos α)−(gsin θ)t2⇒vt1≠vt2

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