Question

A cannonball is catapulted toward a castle. The cannonball’s velocity when it leaves the catapult is 51.6 m/s at an angle of 37.0° with respect to the horizontal, and the cannonball is 7.00 m above the ground at this time. The y-axis points up. What is the y-component of the cannonball’s velocity just before it lands?

Answer 1

Answer:

$v_{y}=35.21m/s$

Explanation:

From the exercise we know the cannonball's initial velocity, the angle which its released with respect to the horizontal and its initial height

$v_{o}=51.6m/s\\\beta =37.0º\\y_{o}=7m$

If we want to know whats the y-component of velocity we need to use the following formula:

$v_{y}^2=v_{oy}^2+ag(y-y_{o})$

Knowing that $g=-9.8m/s^2$

$v_{y}=\sqrt{((51.6m/s)sin(37))^2-2(9.8m/s^2)(0m-7m)}=35.21m/s$

So, the cannonball's y-component of velocity is $v_{y}=35.21m/s$