Question

A bullet is fired from ground level with a speed of 150 m/s at an angle 30.0° above the horizontal at a location where g = 10.0 m/s2. What is the horizontal component of its velocity when it is at the highest point of its trajectory?

Answer 1

Solution: The initial speed of the bullet v = 150 m/s

The angle above the horizontal at which bullet is fired θ = 30^o

Acceleration due to gravity g = 10.0m/s²

There is no gravitational force in the horizontal direction thus no acceleration in horizontal direction. It means that the horizontal component of speed do not change during the entire motion of the bullet and it is same at every point. Thus the horizontal component of speed at the highest point of trajectory is given by

v_x = v*cos(30.0^o)

v_x = (150m/s)*cos(30.0^o)

v_x = 129.9 m/s

Thus the horizontal component of bullet’s speed is 129.9 m/s when it is at the highest point of its trajectory.