Question

If the launch velocity is 115 m/s and the launch angle is 34 degrees, what is the horizontal velocity at the top of the trajectory?

A) 95 m/s
B) 0 m/s
C) 64 m/s
D) 115 m/s

Answer 1

The horizontal velocity at the top of the trajectory for a projectile launched at 115 m/s at an angle of 34 degrees is approximately 95 m/s, corresponding to option A).

Step-by-step explanation:

If the launch velocity is 115 m/s and the launch angle is 34 degrees, the horizontal velocity at the top of the trajectory can be calculated using the cosine component of the launch velocity. Since horizontal velocity remains constant in projectile motion (ignoring air resistance), it can be found using the equation v_x = v × cos(θ), where v_x is the horizontal velocity, v is the launch velocity, and θ is the launch angle.

Using this formula:

v_x = 115 m/s × cos(34°)

This calculation gives us the horizontal velocity as approximately 95 m/s, which corresponds to option A).