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Excellent human jumpers can leap straight up to a height of 110cm off the ground

To reach this height, with what speed would a person need to leave the ground?

How do I solve this question when no other values are given other than distance?

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2 Answers

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Equate potential and kinetic energies: $mgh=\dfrac{mv^2}{2}.$

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When the excellent human jumper is at the highest point, he has a vertical speed $v = 0$ m/s. At the same time, his displacement from ground is $s = 0.11$ m. Throughout the upward motion, he experiences a downward acceleration of $g=9.8$ m/s/s.

Therefore, using the relation between initial speed, final speed, acceleration and displacement,

$$v^2 - u^2 = 2as$$

you should be able to find the initial speed. This is the same as the other answer, which considers kinetic and potential energy.

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