Question

22.A body is thrown with velocity of 40 m/s in a direction making an angle of 30° with the
horizontal. Calculate 1) Horizontal range, 2) Maximum height and 3) Time taken to reach the
maximum height.
Ans. 141.4 m, 20.41 m, 2.041 s​

Answer 1

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Answer 2

Given:-

$u = 40m {s}^{ - 1}$

$\alpha = 30$

(1)

$R = \frac{ {u}^{2} \sin( 2\alpha )}{g}$

$R = \frac{40 \times 40 \times \sin( 2 \times 30)}{9.8}$

$R = \frac{1600 \times \sin(60)}{9.8}$

$R = \frac{1600 \times \sqrt{3} }{2 \times 9.8}$

$R = \frac{800\times \sqrt{3} }{9.8}$

$R = 141.4 \: m$

(2)

$H = \frac{{u}^{2} sin^{2} \alpha\: }{2g}$

$H = \frac{40 \times 40 \times sin^{2} 30 }{2 \times 9.8}$

$H = \frac{1600 \times 1}{{4 \times 2} \times9.8 }$

$H = \frac{200}{9.8}$

$H = 20.41m$

(3)

$T = \frac{2 u \sin( \alpha ) }{2g}$

$T = \frac{2 \times 40 \times \sin(30) }{2 \times 9.8}$

$T = \frac{80 \times 1}{2 \times 2\times 9.8}$

$T = \frac{20}{9.8}$

$T = 2.041s$