Question

A golf ball is hit with an initial velocity of 15 meters per second at an angle of 35 degrees above horizontal. What is the vertical component of the golf ball’s initial velocity ?

Answer 1

Answer:

8.6 m/s

Explanation:

The components of the initial velocity of the ball can be found by using the equations:

$v_x = v cos \theta\\v_y = v sin \theta$

where

$v_x$ is the horizontal component

$v_y$ is the vertical component

v is the magnitude of the velocity

$\theta$ is the angle above the horizontal

Substituting:

v = 15 m/s

$\theta=35^{\circ}$

We find:

$v_x = (15)(cos 35^{\circ})=12.3 m/s\\v_y = (15)(sin 35^{\circ})=8.6 m/s$