A projectile is shot at an angle of 45º to the horizontal near the surface of Earth but in the absence of air resistance.When it reaches the highest point of its trajectory,its speed is 150 m/s.In a second trial with the same projectile,the initial speed is the same but the angle is now 37º with the horizontal.At its highest point in this trajectory,the speed of the projectile would be
A)150 m/s (sin 45º/sin 36º)
B)150 m/s (cos 37º/cos 45º)
C)150 m/s (sin 37º/sin 45º)
D)150 m/s (37/45)
E)None of these is correct.

Question

Quizplus · Accepted Answer

At the highest point in its trajectory, the vertical component of the projectile's velocity is zero, and only the horizontal component remains. The horizontal component of velocity remains constant throughout the flight in the absence of air resistance. The speed at the highest point is given by the initial speed multiplied by the cosine of the launch angle (since cos(θ) gives the horizontal component of the initial velocity vector). Therefore, for the second trial, the speed at the highest point is 150 m/s multiplied by the ratio of the cosines of the new and original launch angles, which is cos(37º)/cos(45º).

A Projectile Is Shot at an Angle of 45º to the Horizontal