Question

During a long jump, an Olympic champion's center of mass rose about 1.2 m from the launch point to the top of the arc. 1) What minimum speed did he need at launch if he was traveling at 6.8 m/s at the top of the arc

Answer 1

Answer: Minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.

Explanation:

Velocity is only in horizontal direction at the top most point which is similar to the velocity in the horizontal direction at the time of launch.

Now, according to the law of conservation of energy the formula used is as follows.

$mgh = \frac{1}{2} mv^{2}_{y}\\v_{y} = \sqrt{2gh}\\= \sqrt{2 \times 9.8 m/s^{2} \times 1.2}\\= 4.85 m/s$

As speed at which the person is travelling was 6.8 m/s. Hence, the initial velocity will be calculated as follows.

$v = \sqrt{v^{2}_{x} + v^{2}_{y}}\\= \sqrt{(6.8)^{2} + (4.85 m/s)^{2}}\\= 11.65 m/s$

Thus, we can conclude that minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.