Physics Question #16

[nc2tarheels]

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https://www.ada.org/oat/oat_sample_test.pdf

It's a smiple kinematics question. It's been 2 yrs since i had physics 1 and this is KILLING me. lol. Anyone know a good place to refresh on kinematics? I've tried Chad's Videos, but to be honest, his physics section is kind of lame. I've also tried khanacademy.org. Thanks

 

[DellGirl]

I can help 🙂. Ok. With projectile motion questions, always separate everything into horizontal and vertical separately.

Horizontal component of velocity is given as 20 m/s. Horizontal distance is given as 50 m. So you can solve for time by using the equation x=vt. So t = x/v = 50/20 = 2.5 seconds for the ball to fall.

Now do the vertical. You want to find change in height. Use the equation y = 1/2at^2. So y = 1/2 * 10 * 2.5^2. Plug 5 * 2.5 * 2.5 into the calculator and you get 31 m.

Note - You are using the same equation for both horizontal and vertical. Change in distance = Vit + 1/2at^2. However, for horizontal, there is no acceleration, so the last part of the equation is 0. For vertical, there is no initial velocity since the ball is thrown horizontally, so the first part of the equation is 0.

I brushed up on my kinematics via Kaplan practice problems, reviewing some of my notes from physics classes, and derekowens videos on youtube.

 

[nc2tarheels]

Thank you so much. I knew I needed to break it up into x and y, just had a brain fart ;0) Thank you!