A 57 O-kg athlete leaps straight up into the air from a trampoline with an initi

Question

A 57 O-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.1 m/s. The goal of this problem is to find the maximum height she and her speed at half maximum height. What are the interacting objects and how do they interact? This answer has not been graded yet Salad the height at which the athlete's speed is 9 1 m/s as y = 0 What is her kinetic energy at this pant? What is the gravitational potential energy associated with the what is her kinetic energy at maximum height? What is the gravitational potential energy associated with the athlete? Write a general equation for energy conservation m this case and serve for the maximum height Substitute and obtain a numerical answer Write the general equation for energy conservation and solve for the velocity at half the maximum height. Substitute and obtain a numerical answer

Explanation / Answer

a] The interacting objects are the trampoline and the athlete where the trampoline provides the athlete enough energy to propel her in the air.

b] y = 0

u = 9.1m/s

therefore, the initial kinetic energy will be: K.Ei = (1/2)mu2 = (1/2)(57)(9.1)2 = 2360.085 J

and at y = 0, the initial potential energy will be: P.Ei = mgy = mg(0) = 0 J

c] At the maximum height, the velocity is zero.

therefore, the kinetic energy at this point will be: K.E = (1/2)m(0)2 = 0 J

by conservation of energy, all the initial kinetic energy will get converted to potential energy at the highest point.

So, the gravitational potential energy at the maximum height will be: P.E = 2360.085 J.

d] mgy = (1/2)mv2

=> 2360.085 = mgy = (57)(9.8)y

=> y = 4.225 m.

e] at half the maximum height, the potential energy = mg(y/2) = 1180.0425 J

therefore, by conservation of energy, the kinetic energy at this point will be: 2360.085 - 1180.0425 = 1180.0425 J

and so, 1180.0425 = (1/2)mv2 = (1/2)(57)v2

=> v = 6.435 m/s. This is the velocity at half the maximum height.

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