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(C) The horizontal velocity $u_x = u \cos 	heta$ is equal to the velocity acquired from rest with acceleration $a = 3 \, ms^{-2}$ for time $t = 0.5 \

Vedclass · Accepted Answer

$	an^{-1}(4/9)$

Vedclass · Answer

(C) The horizontal velocity $u_x = u \cos 	heta$ is equal to the velocity acquired from rest with acceleration $a = 3 \, ms^{-2}$ for time $t = 0.5 \, minutes = 30 \, s$.$u_x = a 	imes t = 3 	imes 30 = 90 \, ms^{-1}$.The maximum height $H$ is given by $H = \frac{u_y^2}{2g}$,where $u_y = u \sin 	heta$.$80 = \frac{(u \sin 	heta)^2}{2 	imes 10} \implies (u \sin 	heta)^2 = 1600 \implies u_y = 40 \, ms^{-1}$.The angle of projection $	heta$ is given by $	an 	heta = \frac{u_y}{u_x}$.$	an 	heta = \frac{40}{90} = \frac{4}{9}$.Therefore,$	heta = 	an^{-1}(4/9)$.

$A$ projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $3 \, ms^{-2}$ for $0.5 \, minutes$. If the maximum height reached by it is $80 \, m$,then the angle of projection is (Take $g = 10 \, ms^{-2}$)

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