In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole. With what minimum speed must the athlete leave the ground in order to lift his center of mass 2.35 m and cross the bar with a speed of 0.65 m/s?
2007-02-16 08:19:35 · 3 answers · asked by Jessie L 2 in Science & Mathematics ➔ Physics
This one requires some thought. At the instant the athlete is over the bar, his/her vertical velocity is zero. Only remains the horizontal component of velocity, which is given in the statement, 0.65 m/s.
Horizontal motion, NOT being accelerated motion, is therefore constant. You can assume that the initial value of horizontal velocity is the same, namely, 0.65 m/s. In order to solve this problem, you only need to find the vertical component of the initial velocity, which is being asked.
It can be found form kinematic formulas, or from energy considerations. At the highest point of trajectory, potential energy is mgh. When leaving the ground, there's kinetic energy only, ½ mv². Equating,
½ mv² = mgh,
or v² = 2gh = 46.11 m²/ s²
v = 6.79 m/s
We're not over, yet. This is the vertical component only. I'm sure you know how to obtain resultant velocity. It's not very different from the vertical component, 6.821∠84.53° m/s.
By the way, whether energy is regarded as positive or negative is purely a matter of convention. In one of your previously posted problems you insist that energy is negative, because work has been done on the car. You may as well consider it positive, since the car is returning the kinetic energy it had. Theoretically, at least, this energy could be used to do work.
2007-02-16 09:12:15 · answer #1 · answered by Jicotillo 6 · 0⤊ 1⤋
You need to look up the equation to calculate the max height of a jump in the vertical direction... you've got the max height, so solve for the initial velocity.
that gives you the speed in the vertical direction. We're going to assume that they want the 0.65 m/s to be in the horizontal, so that's your velocity in the horizontal.
Use good old pythagorus to solve: A squared + B squared = C squared where C will be the minimum velocity the athlete must leave the ground at
2007-02-16 17:06:15 · answer #2 · answered by promethius9594 6 · 0⤊ 0⤋
go read your textbook and think some. thinking is good.
googling is bad.
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long time since i did such stuff.
you will get 2 simultaneous equations - one for the horizontal component, other for vertical. solve them.
2007-02-16 16:29:12 · answer #3 · answered by Anonymous · 1⤊ 1⤋