Question

If you throw a ball at 35° and 10 m/s what is the horizontal component of velocity

Answer 1

Answer:

8.2 m/s

Explanation:

The horizontal component of the velocity is given by:

$v_x = v cos \theta$

where

v = 10 m/s is the magnitude of the velocity

$\theta=35^{\circ}$ is the angle at which the ball has been thrown, with respect to the horizontal

Substituting the values into the equation, we get

$v_x=(10 m/s)(cos 35^{\circ})=8.2 m/s$