Question

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

0.0m/s
10m/s
5.7m/s
8.2m/s

Answer 1

8.2 m/s

Step-by-step 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`