Question

If R = 5T^2. find out the angle of projection.

A 45°
B. 60°
C. 90°
D. 15°​

Answer 1

Answer:

Range of projectile -

$R = \frac{ {u}^{2} sin2θ}{g} ......(1)$

Time of flight -

$T = \frac{2u \sinθ }{g} .......(2)$

initial speed - u

gravity (g)- 10 m/s²

R=5T² ......... (3)

now, substituating equation (3) by equation (1) and (2) .

$= > \frac{ {u}^{2} \sin2θ }{g} = 5( { \frac{2u \sinθ}{g} )}^{2} \\ = > \frac{ {u}^{2} \: 2sinθ \: cosθ}{g} = 5( \frac{4 {u}^{2} {sin}^{2}θ }{ {g}^{2} }) \\ = > \frac{ {u}^{2} \: 2sinθ \: cosθ}{g} = \frac{20 {u}^{2} {sin}^{2}θ }{ {g}^{2} } \\ = > cosθ \: = \frac{10sinθ}{g} \\ = > \frac{sinθ}{cosθ} = \frac{g}{10} \\ = > tan = \frac{10}{10} \\ = > tanθ = 1 \\ = > θ = \frac{\pi}{4} \\ = > θ = 45°$

Hence, angle of projectile is A. 45° .