Question

$\large\rm{\underline{Question:-}}$

If a horizontal range of projectile be a and the maximum height attained by it is b , then prove that the velocity of projection is :
$\sf{\biggl\{2g\left(b+\dfrac{a^2}{16b}\right) \biggl\}^{\frac{1}{2}}}$
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Answer 1

Explanation:

Solution

a=R=gv2sin2θ⟶1

b=H=2gv2sin2θ⟶2

From equation 1 and 2

ab=tanθ×41t

tanθ=a4b,sinθ=16b2+a24b

Considering equation 2

b=2gv2×(16b2+a2)16b2

v2=8g(16b2+a2)

v=[2g(b2+16a2)]21

Answer 2

Answer:

Refer to the above attachment for the required solution.

Hope it helps you.

✌️Santa19 ✌️