Question

please answer this question as quick as possible​

Answer 1

$\large{ \underline{ \mathcal{Solution :}}}$

Total time of motion is given by

$t = \frac{2 v_{0} sin \alpha }{g}$

$sin \alpha = \frac{tg}{2 v_{0} }$

$sin \alpha = \frac{9.8t}{2 \times 240}$

.......(1)

and

Horizontal range

$R = v_{0}cos \alpha t$

$cos \alpha = \frac{R}{ v_{0}t }$

$cos \alpha = \frac{1500}{240t} = \frac{85}{4t}$

........(2)

From equations (1) & (2)

$\frac{ {(9.8)}^{2} {t}^{2} }{ {(480)}^{2} } + \frac{ {(85)}^{2} }{ {(4 {t}^{2} )}^{2} } = 1$

On simplifying

${t}^{4} - 2400 {t}^{2} + 1083750 = 0$

Solving for t^2, we get

${t}^{2} = \frac{2400 + - \sqrt{1425000} }{2} = \frac{2400 + - 1194}{2}$

Thus

t = 42.39s = 0.71 min and

t = 24.55s = 0.41 min depending on the angle /alpha

Answer 2

Answer:

$Solution:</p><p></p><p>Total time of motion is given by</p><p></p><p>t = \frac{2 v_{0} sin \alpha }{g}t=g2v0sinα</p><p></p><p>sin \alpha = \frac{tg}{2 v_{0} }sinα=2v0tg</p><p></p><p>sin \alpha = \frac{9.8t}{2 \times 240}sinα=2×2409.8t</p><p></p><p>.......(1)</p><p></p><p>and</p><p></p><p>Horizontal range</p><p></p><p>R = v_{0}cos \alpha tR=v0cosαt</p><p></p><p>cos \alpha = \frac{R}{ v_{0}t }cosα=v0tR</p><p></p><p>cos \alpha = \frac{1500}{240t} = \frac{85}{4t}cosα=240t1500=4t85</p><p></p><p>........(2)</p><p></p><p>From equations (1) & (2)</p><p></p><p>\frac{ {(9.8)}^{2} {t}^{2} }{ {(480)}^{2} } + \frac{ {(85)}^{2} }{ {(4 {t}^{2} )}^{2} } = 1(480)2(9.8)2t2+(4t2)2(85)2=1</p><p></p><p>On simplifying</p><p></p><p>{t}^{4} - 2400 {t}^{2} + 1083750 = 0t4−2400t2+1083750=0</p><p></p><p>Solving for t^2, we get</p><p></p><p>{t}^{2} = \frac{2400 + - \sqrt{1425000} }{2} = \frac{2400 + - 1194}{2}t2=22400+−1425000=22400+−1194</p><p></p><p>Thus</p><p></p><p>t = 42.39s = 0.71 min and</p><p></p><p>t = 24.55s = 0.4</p><p></p><p>$