Question

A bomb is fired from a cannon with a velocity of of $1000 \,m/s$ making an angle of $ 30^{\circ} $ with the horizontal. The time taken by the bomb to reach the highest point, will be:

• A. 51 sec
• B. 25.5 sec
• C. 61 sec
• D. 21 sec

Answer 1

Answer: (A)

Here: Velocity of bomb $v=1000 \,m / s$
The angle $\theta=30^{\circ}$
Time taken by the bomb to reach the highest point
$t=\frac{v \sin \theta}{g} $
$=\frac{1000 \sin 30^{\circ}}{9.8} $
So, $ t=\frac{1000 \times 0.5}{9.8}=51\, \sec$