An artilleryman makes a training shot from a gun at an angle of 60 degrees to the horizon.

An artilleryman makes a training shot from a gun at an angle of 60 degrees to the horizon. The initial velocity of the projectile is 500 m / s. How long after the shot will the artilleryman hear the sound from the bursting of the shell?

Reference system: origin is at the point of the shot, y-axis is directed upwards, x-axis is horizontal.

Vertical and horizontal initial velocities:

Vy0 = V0 * cos 60 ° = (500 * √3) / 2 = 250√3 m / s;

Vx0 = V0 * sin 60 ° = 500 * 0.5 = 250 m / s.

The projectile will move vertically with an acceleration g = 10 m / s ^ 2. The horizontal component of the speed is constant.

During the time t1 in flight, the vertical movement of the projectile is zero.

Moving in y through Vy0 and accelerating g:

0 = vy0t1 – gt1 ^ 2/2

vy0 = gt1 / 2;

t1 = 2vy0 / g.

Answer: 150.28 c.