# Determine the magnitude and direction of the velocity at which the projectile is projected.Two seconds after projection, a projectile is travelling in a direction inclined at an angle 30 degrees to...

Determine the magnitude and direction of the velocity at which the projectile is projected.

Two seconds after projection, a projectile is travelling in a direction inclined at an angle 30 degrees to the horizontal. After one more second it is travelling horizontally.

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### 1 Answer

Two seconds after being launched the direction of the projectile's velocity is at an angle 30 degrees to the horizontal. After 3 seconds it is traveling parallel to the horizontal. The magnitude and angle at which the projectile is projected has to be determined.

Let the magnitude at which it is launched be V and the angle with the horizontal be A. There is an acceleration on the projectile of 9.8 m/s^2 acting vertically downwards. In the horizontal direction there is no acceleration. When the projectile is launched the component of the vertical velocity is V*sin A and the component of velocity in the horizontal direction is V*cos A.

As the projectile travels horizontally after 3 seconds, its vertical velocity is 0.

=> 0 = V*sin A - 9.8*3

=> V*sin A = 29.4...(1)

After 2 seconds, the vertical velocity is V*sin A - 9.8*2 and the horizontal velocity is V*cos A. As the angle of the velocity is 30 degrees to the horizontal tan 30 = (V*sin A - 2*9.8)/V*cos A

=> tan 30 = 9.8/V*cos A

=> V*cos A = 16.97...(2)

From (1) and (2)

tan A = 29.4/16.97

=> A = arc tan (29.4/16.97) = 60 degrees

V = 29.4/sin 60 = 33.94 m/s

**The projectile is launched at 33.94 m/s at an angle 60 degrees to the horizontal.**

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