Question

A bomber flying upward at an angle of 53${}^{\circ }$ with the vertical releases a bomb at an altitude of 800 m. The bomb strikes the ground 20 s after its release. If g = 10 m s${}^{-2}$,the velocity at the time of release of the bomb in $m{s}^{-1}$ is

• A. 400
• B. 800
• C. 100
• D. 200

Answer 1

Answer: (C)

-h - ut - $\frac{1}{2}$ g t${}^2$