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Question

(A) The horizontal component of the velocity of the bomb remains constant throughout the motion because there is no acceleration in the horizontal dir

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The horizontal component of the velocity of the bomb remains constant throughout the motion because there is no acceleration in the horizontal direction.$V_x = u = 500 \ m/s$The vertical component of the velocity at time $t = 10 \ s$ can be calculated using the first equation of motion:$V_y = u_y + g t$Since the bomb is released horizontally,the initial vertical velocity $u_y = 0$.$V_y = 0 + (10 \ m/s^2) 	imes (10 \ s) = 100 \ m/s$The angle $	heta$ that the velocity vector makes with the horizontal is given by:$	an 	heta = \frac{V_y}{V_x}$$	an 	heta = \frac{100}{500} = \frac{1}{5}$$	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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