A bomb is fixed from a canon with a velocity of 1000 m/s making an angle of 30° with the horizontal (g=9.8 text m/s2) . Time taken by bomb to reach the highest point is:

Question

A bomb is fixed from a canon with a velocity of 1000 m/s making an angle of 30° with the horizontal (g=9.8 text m/s2) . Time taken by bomb to reach the highest point is: (A) 40 s (B) 30 s (C) 51 s (D) 25 s

Tardigrade Admin · Accepted Answer

T=(u sin θ /g) Given, u=1000 text m/s, θ =30° ,g=9.8m/s2 T=(1000× sin 30° /9.8)=51 s

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

40 s

30 s

51 s

25 s

15s

$T = \frac{u s i n θ}{g}$ Given, $u = 1000 m / s,$ $θ = 30^{\circ}, g = 9.8 m / s^{2}$ $T = \frac{1000 \times s i n 30 ^{\circ}}{9.8} = 51 s$

Q. A bomb is fixed from a canon with a velocity of 1000 m/s making an angle of 30∘ with the horizontal (g=9.8m/s2) . Time taken by bomb to reach the highest point is: