Question

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

• A. ${tan}^{-1}$($\frac{1}{5}$)
• B. tan ($\frac{1}{5}$)
• C. ${tan}^{-1}$(1)
• D. ${tan}^{-1}$(5)

Answer 1

The correct option is A ${tan}^{-1}$($\frac{1}{5}$)

Horizontal component of velocity $v_x$ = 500 m/s
and vertical components of velocity while striking the ground.
$v_y$ = 0 + 10 $\times$ 10 = 100 m/s
$\therefore$ Angle with which it strikes the ground.
$\theta$ = ${tan}^{-1}$($\frac{v_y}{v_x}$) = ${tan}^{-1}$($\frac{100}{500}$)