Derive :
[tex] h_{max} = \dfrac{u^{2}}{2g}[/tex]
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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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[tex]h_{max} = \dfrac{u^{2}}{2g}[/tex]
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ux=ucosθax=0
uy=usinθay=−g
atmax.height
Vg=0
⇒0−usinθ=gt
[tex]⇒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} [/tex]