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How do you calculate the initial vertical velocity of this projectile motion (lab)?...

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seventhheaven | Student, Undergraduate | (Level 1) Honors

Posted October 2, 2012 at 7:06 AM via web

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How do you calculate the initial vertical velocity of this projectile motion (lab)? Please help!

So basically in the lab, we threw a tennis ball into a field and measured the initial height it's thrown at (dy), distance it travelled (dx), and time (s).

I recorded all the data:

dy = 1.41 m

dx = 13 m

t = 3.6 s

The projectile looks like this (link): http://www.real-world-physics-problems.com/images/projectile_motion_1.png

I'm very confused because there's an initial height to this problem. I calculated the horizontal velocity, but I can't figure out how to calculate the (initial) vertical velocity.

Oh, and please explain how you did it! I really wanna understand it. Thanks in advance!

More info (this is the actual lab): https://docs.google.com/viewer?a=v&q=cache:aH6of_Q11ckJ:prettygoodphysics.wikispaces.com/file/view/Projectile%2BMotion%2BLab.pdf+&hl=en&gl=ca&pid=bl&srcid=ADGEESjq0plBrJoYCppCUqc7iUf35p5gT0JmPSvQC1Qmv6OTZ9JSp3cgq5OR9s3WnwK9x6pNSPgOcVhGNaP8PPk04Vz3ETxkj_oodZIBqXoq2SX0i5Ga-3n4W8ch7mFFNkz7beVvcWes&sig=AHIEtbS4tM4j7Nk08aNmcn9XkWcyamDDFg

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted October 2, 2012 at 12:37 PM (Answer #1)

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Consider the whole motion of the ball from time t = 0 to t = 3.6 to the vertical direction. When you consider the initial position it is point 1 as in your link and final position is 2.

Consider we are measuring the vertical height upwards. lets take point 1 as vertical height h = 0 and point 2 as vertical height h=-1.41m.

Now lets put the velocity equation `S = ut+1/2*a*t^2` to the vertical direction for the whole motion.

u is the vertical component that you want to find.

S = -1.4 because s is the vertical height difference between initial position and final position.

t = 3.6

a = -g because the ball runs under gravity.

 

`uarr S = ut+1/2*a*t^2`

`-1.4 = u*3.6+1/2*(-9.81)*3.6^2`

    u = 17.27m/s

 

So initial vertical velocity of the ball is 17.27m/s.

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