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Alenkasestr [34]
3 years ago
5

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?
Physics
1 answer:
Andrei [34K]3 years ago
6 0

Answer:

v_{y}=35.21m/s

Explanation:

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

v_{o}=51.6m/s\\\beta =37.0º\\y_{o}=7m

If we want to know whats the <u>y-component of velocity</u> 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

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