Question

Derive :

$h_{max} = \dfrac{u^{2}}{2g}$



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Answer 1

Answer:

v^2=u^2+2gh

0=u^2+2gh

Gravitational acceleration will be negative because it is against the gravity.

0=u^2+2×-gh

0=u^2-2gh

2gh=u^2

h=u^2/2g. Hence proved.

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Answer 2

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$h_{max} = \dfrac{u^{2}}{2g}$

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ux=ucosθax=0

uy=usinθay=−g

atmax.height

Vg=0

⇒0−usinθ=gt

$⇒t= \frac{usinθ}{g} \\ ∴H=ugt− \frac{1}{2} ayt 2 \\ ⇒H= \frac{usinθ}{g} .usinθ− \frac{1}{2} \times g \times \frac{u²sin²0}{g2} \\ ∴H= \frac{usin0}{g} .usin0 \frac{1}{2} \times g \times \frac{u²sin²0}{g2} \\ ∴H= \frac{u²sin²0}{2g} \: when ball comes to gound t = \frac{du sin 0}{g} \\ ∴Range = Ux.T + \frac{1}{2} a × T² \\ = U cos 0 = \frac{2u sin 0}{g} \\ ∴Range = \frac{u²sin²0}{g}$