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

Question

A bomb is fired from a cannon with a velocity of of 1000 m/s making an angle of 30° 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

Tardigrade Admin · Accepted Answer

Here: Velocity of bomb v=1000 m / s The angle θ=30° Time taken by the bomb to reach the highest point t=(v sin θ/g) =(1000 sin 30°/9.8) So, t=(1000 × 0.5/9.8)=51 sec

Q. A bomb is fired from a cannon with a velocity of of $1000 m / s$ making an angle of $3 0^{\circ}$ with the horizontal. The time taken by the bomb to reach the highest point, will be:

51 sec

25.5 sec

61 sec

21 sec

Here: Velocity of bomb $v = 1000 m / s$
The angle $θ = 3 0^{\circ}$
Time taken by the bomb to reach the highest point
$t = \frac{v s i n θ}{g}$
$= \frac{1000 s i n 3 0 ^{\circ}}{9.8}$
So, $t = \frac{1000 \times 0.5}{9.8} = 51 sec$

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