Question

A body is projected with initial velocity of 40 m/s and at an angle of 30°with the horizontal. Calculate (i) time of flight and
(ii) horizontal range of projectile.(assume g= 10 m/s^2)​​

Answer 1

Explanation:

Time of Flight (t)

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

$t \: = \frac{2 \times 40 \: \sin(30) }{10}$

$t \: = \frac{2 \times 40 \times \frac{1}{2} }{10}$

$t \: = 4 \: sec$

Horizontal Range (R)

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

$R \: = \frac{ {(40)}^{2} \sin(2 \times 30) }{10}$

$R = \frac{6400 \times \sin(60) }{10}$

$R \: = \frac{6400 \times \frac{ \sqrt{3} }{2} }{10}$

$R = 554.25 \: m$

Answer 2

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