Question

A short is fired with a velocity 100m/s in a direction making an angle 60°with the vertical calculate its time of flight the maximum height reached and its horizontal range?

Answer 1

Hope it helps.......................

Answer 2

Time of flight$=\frac{2usin\theta}{g}$

$=\frac{2\times 100\times sin 30}{10}$

$=20\times \frac{1}{2}$

$=10sec$

Max height $=\frac{u^2sin^2\theta}{2g}$

$=\frac{100\times 100}{2\imes 10}\times \frac{1}{2}\times \frac{1}{2}$

$=125M$

Horizontal range $=\frac{u^2 sin2\theta}{g}$

$=\frac{100\times 100\times sin 60}{g}$

$=\frac{100\times 10 }{10} \times \frac{\sqrt{3} }{2}$

$=500\sqrt{3} M$

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