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Question

(A) The horizontal component of velocity remains constant throughout the motion: $v_x = 500\,m/s$.The vertical component of velocity when the bomb str

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{1}{5}ight)$

Vedclass · Answer

(A) The horizontal component of velocity remains constant throughout the motion: $v_x = 500\,m/s$.The vertical component of velocity when the bomb strikes the ground is given by $v_y = u_y + gt$. Since the initial vertical velocity $u_y = 0$,we have:$v_y = 0 + 10\,m/s^2 	imes 10\,s = 100\,m/s$.The angle $	heta$ that the velocity vector makes with the horizontal when it strikes the ground is given by:$	an	heta = \frac{v_y}{v_x} = \frac{100}{500} = \frac{1}{5}$.Therefore,$	heta = 	an^{-1}\left(\frac{1}{5}ight)$.

$A$ bomber plane moves horizontally with a speed of $500\,m/s$ and a bomb released from it,strikes the ground in $10\,s$. The angle with the horizontal at which it strikes the ground will be $(g = 10\,m/s^2)$.

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