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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
1 Answer | Add Yours
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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