The athlete runs up and jumps in length. Takeoff and push time 4 s. The coefficient of friction of the soles on the treadmill

The athlete runs up and jumps in length. Takeoff and push time 4 s. The coefficient of friction of the soles on the treadmill is 0.15. The maximum jump height is 0.8 m. Determine the length (in m) of the jump. The sliding friction force during the takeoff run and push is considered maximum and acting continuously. Neglect air resistance. Free fall acceleration 10 m / s2. The answer is a whole number, if necessary, round to whole numbers.

The athlete runs up and jumps in length. The resultant force F, acting on the athlete, with mass m, can be found according to Newton’s second law F = m ∙ a, where a is the acceleration of the run. The vertical forces cancel each other out. In the horizontal direction the athlete has to overcome the friction force: Ffr = μ ∙ m ∙ g, where the acceleration of gravity is g ≈ 10 m / s². Then m ∙ a = μ ∙ m ∙ g or a = μ ∙ g. The horizontal speed acquired by the athlete during the takeoff run tp by the time of the jump will be v = vо + a ∙ tр, where the initial speed of the takeoff run is vо = 0 m / s. We get v = μ ∙ g ∙ tr. When jumping, the athlete reached the height h in t seconds, where h = g ∙ t² / 2, then the time of the entire jump tpr will be tpr = 2 ∙ (2 ∙ h / g) ^ (1/2), while the flight range L = v ∙ tpr; L = μ ∙ g ∙ tр ∙ 2 ∙ (2 ∙ h / g) ^ (1/2) since the air resistance can be neglected. From the condition of the problem it is known that the take-off time tр = 4 seconds, the coefficient of friction of the soles on the treadmill is 0.15, the maximum jump height is 0.8 m.We substitute the values ​​of physical quantities in the calculation formula and make calculations: L = 0.15 ∙ 10 ∙ 4 ∙ 2 ∙ (2 ∙ 0.8 / 10) ^ (1/2) = 4.8 ≈ 5 (m).
Answer: the jump length is ≈ 5 meters.