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Home Q & A Board Science-Related Homework Help Physics
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ripple ripple
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10 years ago
A kangaroo can jump a height of 2.62m.Determine the take-off speed of the kangaroo...
I don't need the solutions but why must you take the final velocity when it is at the highest point but not lowest point when it has stopped jumping?
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#1
10 years ago
The equation for motion with constant acceleration:

V^2 = U^2 + 2 x a x s

where V is the final velocity and U is the  starting velocity, applies only whilst the direction of the motion does not reverse (after becoming zero).  In this case the "final velocity" becomes zero when the motion reaches the highest point in the trajectory.  

s is distance (height), whilst a is the acceleration (in this case of gravity).

For motion where the direction is unchanged the equation applies when V and U are starting and final velocities, however in this particular case the direction does change and one should treat the motion as two separate ones which are connected where their directions reverse.

The final or the starting velocities can be interchanged, providing the sign of the acceleration is allowed to change too so as to result in positive accelerations due to increases in the magnitude of the velocities or V > U .
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tpwatkintpwatkin
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10 years ago
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