Question

A bomb is fixed from a canon with a velocity of $1000m/s$ making an angle of ${30}^o$ with the horizontal $g=9.8m/s^2$. Time taken by bomb to reach the highest point is

• A. $40s$
• B. $30s$
• C. $51s$
• D. $25s$

Answer 1

Answer: (C)

$51s$

For a projectile motion, if u be the initial velocity of the bomb and $\theta$ be the angle of projection, the total time of flight will be $T=\frac{2u\sin \theta }{g}$

The total time of flight is twice the time $t$ taken by bomb to reach highest point.

i.e $T=2t$

or $t=T/2=\frac{u\sin \theta }{g}=\frac{1000\sin 30}{9.8}=51s$