Question

A motorcycle stunt person leaves the jump ramp traveling at 35 m/s at 22° above the horizontal

Answer 1

Missing questions (found on google):

A. Draw a vector representing the velocity of the motorcycle

B. Draw the horizontal and vertical components of the vector

C. Use vector resolution to find the magnitude of each component

Solution:

A-B) I have attached a picture including both the vector and its components.

Every vector can be represented using an arrow, whose length is proportional to its magnitude, and the direction is shown by the arrow.

The velocity of the motorbike here has magnitude of 35 m/s at an angle $22^{\circ}$ above the horizontal, so it is represented as vector v in the picture, with

$|v|=35 m/s\\\theta = 22^{\circ}$

The horizontal and vertical components are also represented, obtained by projecting the vector itself along the x- and y- axis, and they are indicated with $v_x, v_y$, respectively.

C) 32.5 m/s, 13.1 m/s

The magnitude of each component can be found using the following equations:

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

where

v = 35 m/s is the magnitude of the vector

$\theta=22^{\circ}$ is the angle of the projection

Substituting into the equations, we find

$v_x = (35)(cos 22)=32.5 m/s$

$v_y = v sin \theta = (35)(sin 22)=13.1 m/s$