Physics Help Forum initial horizontal and vertical velocity problem

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Mar 31st 2011, 02:30 AM #1 Junior Member   Join Date: Mar 2011 Posts: 2 initial horizontal and vertical velocity problem hi I have just started a new uni course and physics is one of my subjects as part of the course. This is my first time learning the subject so if the problem is pretty bad, ah well! I am really stuck on this problem.. Jai leaves the long jump take off board with a vertical velocity of 4m/s and a horizontal velocity of 8m/s. If Jai's centre of gravity is 1.5m high at the instant of take off, how high will he be at his peak of flight? now the way i tackled this was that i wasn't to sure which options to use. I didn't know which was right as i get stuck on the differences. The first way was use the initial velocity that we were give and find the angle and side: Cos (theta) = 4/8 = 60 deg then height (x) = sin 60 x 4 = 3.46 m OR do we use the 4m/s as the initial velocity and then find the displacement as it would represent the height! vf^2= Vi + 2as vf is 0 as peak you change direction 0= 4 + 2(-10)s 0= 4-20s s=16 metres So i am not to sure, please help!!
Mar 31st 2011, 07:18 AM   #2

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 Originally Posted by boubba_25 hi I have just started a new uni course and physics is one of my subjects as part of the course. This is my first time learning the subject so if the problem is pretty bad, ah well! I am really stuck on this problem.. Jai leaves the long jump take off board with a vertical velocity of 4m/s and a horizontal velocity of 8m/s. If Jai's centre of gravity is 1.5m high at the instant of take off, how high will he be at his peak of flight? now the way i tackled this was that i wasn't to sure which options to use. I didn't know which was right as i get stuck on the differences. The first way was use the initial velocity that we were give and find the angle and side: Cos (theta) = 4/8 = 60 deg then height (x) = sin 60 x 4 = 3.46 m
But then max height has the units of m/s, not m.

 Originally Posted by boubba_25 OR do we use the 4m/s as the initial velocity and then find the displacement as it would represent the height! vf^2= Vi + 2as
The equation is vf^2 = vi^2 + 2as.

Since motion in the x and y directions are independent of each other all you need is the vertical component of velocity at the start and then at your maximum height. You have the vertical initial component of velocity, so to find the max height all you need to do is set vfy = 0 m/s. Note carefully that the overall velocity is NOT 0 m/s at maximum height. Jai still has a horizontal component of velocity there.

-Dan
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 Apr 1st 2011, 07:02 PM #3 Junior Member   Join Date: Mar 2011 Posts: 2 okayyy, sweet thanks for your help! I understand where to go from here now!

 Tags horizontal, initial, problem, velocity, vertical

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