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A ball thrown at different angles with the same speed u and from the same point and it has the same range in both cases. If y1 and y2 are the heights attained in two cases, then y1 + y2 is equal to

u^2/g

2u^2/g

u^2/2g

u^2/4g

Prathamesh Jagtap , 9 Years ago

Grade 11

1 Answers

Raman Mishra

Last Activity: 9 Years ago

Two projectiles have the same range if they are launched at angles which are complementary to each other. That is, if the 1^st projectile is launched at an angle $\theta$ from the horizontal, then the other projectile must be launched at an angle of $\frac{\pi }{2} -\theta$ from the horizontal for having the same range.

Now, maximum height of a projectile = $h = \frac{u^{2}\sin ^{2}\theta}{2g}$

For the 1^st ball we have $h_{1} = \frac{u^{2}\sin ^{2}\theta}{2g}$

For the 2^nd ball, we have $h_{2} = \frac{u^{2}\sin ^{2}(\frac{\pi }{2}-\theta )}{2g}$ $= \frac{u^{2}\cos ^{2}\theta }{2g}$

$\therefore$ h₁ + h₂ = u²/2g

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