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How do you calculate the initial vertical velocity of this projectile motion (lab)?...
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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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.
Posted by jeew-m on October 2, 2012 at 12:37 PM (Answer #1)
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