a 950n basketball player jumps to block a jump shot. here is what you know: the average force applied to the floor: 1800n force is applied over a vertical distance of .5m (from the lowest point in the squat to the position at the instant of takeoff.

It is intimately correlated with both the amount of force applied to the item and its motion. The equation below may be used to compute work: Work is equal to force times distance. The Newton meter is the SI unit of work (N m). When an item is moved across a distance of 1 m with 1 N of force, one joule is equal to the amount of work that is accomplished.
The forces and motion in a standing vertical leap are both vertical. F = FGRF - mg, where FGRF is the vertical ground reaction force applied to the jumper and mg is the jumper's body weight, results in a vertical force on the jumper.

Given :

Weight of the Basket ball Player = 950 N.

Force Applied on the Floor = 1800 N

over Vertical distance = .5 m

resultant force f = 1800 N - 950 N = 850 N

work done = 850 * 0.5 m =425 J

kenetic energy = 1/2 mv^2= 425

mass = 950/ g = 96.9

v^2 = 425 * 2/ 96.9

v= [tex]\sqrt{8.78}[/tex] = 2.96 m/s

instant Take off is 2.96 m/s

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