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Solve the following :

Question:

A ball is thrown at a speed of $40 \mathrm{~m} / \mathrm{s}$ at an angle of $60^{\circ}$ with the horizontal.

Find

(a) The maximum height reached

(c) The range of the ball

Take $\mathrm{q}=10 \mathrm{~m} / \mathrm{s}^{2}$

Solution:

$\mathrm{u}=40 \mathrm{~m} / \mathrm{s} ; \theta=60^{\circ}$

(a)

$\max ^{=} \frac{\mathrm{u}^{2} \sin 2 \theta}{2 \mathrm{~g}}=\frac{40^{2}\left(\sin ^{2} 60\right)}{2 \mathrm{~g}}$

$\mathrm{H}_{\max }=60 \mathrm{~m}$

(b) 

$\mathrm{R}=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}=\frac{40^{2} \sin (2 \mathrm{x} 60)}{\mathrm{g}}$

$R=80 \sqrt{3} \mathrm{~m}$

 

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