The artilleryman heard the sound of a shell bursting at an angle of 450 to the horizon

The artilleryman heard the sound of a shell bursting at an angle of 450 to the horizon, 2 minutes 30 seconds after the shot. What was the muzzle velocity of the projectile?

t = 2 min 30 s = 150 s.

∠α = 45 °.

g = 9.8 m / s2.

V0 -?

The projectile moves horizontally uniformly with a speed Vg = V0 * cosα.

Vertically, the projectile first moves equally decelerated to a complete stop, and then uniformly accelerated with the acceleration of gravity g.

The projectile flight time t is expressed by the formula: t = 2 * V0 * sinα / g.

The initial velocity of the projectile V0 will be determined by the formula: V0 = g * t / 2 * sinα.

V0 = 9.8 m / s2 * 150 s / 2 * 0.7 = 1050 m / s.

Answer: the speed of the projectile at the moment of firing is V0 = 1050 m / s.