After one second the velocity of a projectile makes an angle of $45{^\circ}$ with the horizontal. After another one second it is travelling horizontally. The magnitude of its initial velocity and angle of projection are $(g = 10\ ms^{- 2})$

Question

Dr. Abubacker · Accepted Answer

(b) Time of flight of this particle, $T = 4$ s. if $u$ is its initial speed and $	heta$ is the angle of projection, then
$T = 4 = \frac{2u\sin	heta}{g}$ or $u\sin{	heta = 2g}$ …(i)
After 1 s, the velocity vector of particle makes an angle of $45{^\circ}$ with horizontal, so
$v_{x} = v_{y}\ ie,u\cos{	heta = \left( u\sin	heta ight) - gt}$
or $u\cos{	heta = 2g - g}\ \ \ \ \ (	herefore t = 1s)$
or $u\cos{	heta = g}$ …(ii)
Squaring and adding Eqs. (i) and (ii), we have
$u^{2} = 5g^{2} = 5(10)^{2} = 500$
or $u = \sqrt{500} = 22.36\ ms^{- 1}$
Dividing Eq. (i) by Eq. (ii), we have
$	an	heta = 2\ \ or\ \ 	heta = 	an^{- 1}(2)$

Q. After one second the velocity of a projectile makes an angle of $45{^\circ}$ with the horizontal. After another one second it is travelling horizontally. The magnitude of its initial velocity and angle of projection are $(g = 10\ ms^{- 2})$

Explanation

Reviewed by Dr. Abubacker

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