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!

### 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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