a 62.5 kg athlete leaps straight up into the air from a trampoline with an initi

Question

a 62.5 kg athlete leaps straight up into the air from a trampoline with an initial speed of 7.1 m/s. the goal of this problem is to find the A 62.5-kg athlete leaps straight up into the air from a trampoline with an initial speed of 7.i m/s. The goal of this problem is to find the maximum height she attains and her speed at half maximum height (a) What are the interacting objects and how do they interact? This answer has not been graded yet (b) Select the height at which the athlete's' speed is 7.1 m/s as y - 0. What is her kinetic energy at this point? What is the graveational potential energy associated with the athlete? (e) what is her kinetic energy at maximum height? What is the gravitational potential energy associated with the athlete? (d) Write a general equation for energy conservation in this case and solve for the maximum height. Substitute and obtain a numerical answer (e) Write the general equation for energy conservation and solve for the velocity at haif the maximum height. Substitute and obtain a numerical answer mýs

Explanation / Answer

Given : m=62.5 kg ; u=7.1 m/s

Solution :

(a) Interacting object is athlete , trampoline and air if we consider air resistance

(b)

Using the equation :

v2= u2+2as

When person reaches the maximum height, the velocity at that point would be 0.

So 0= (7.1)2 -2(9.81)s [a = -g , upward motion]

s = (7.1)2/(2x9.81)

= 2.57 m [ maximum height ]

Half height would be 2.57/2= 1.285 m

Again using the equation v2 = u2+2as

v2= (7.1)2- 2(9.81)(1.285)

v = 5.02 m/s

Since they define this height as y=0 hence potential energy would be zero relative to this point.

At this point person has all kE but no PE.

KE = (1/2)mv2

= (1/2)(62.5)(7.1)2

=1575.31 J

(c) At the maximum height , her velocity is 0 hence kinetic energy would be 0.

Potential energy PE = mgh = (62.5)(9.81)(2.57) = 1575.73 J

(d) By conservation of energy

Initial energy = Final energy

KEi + PEi= KEf + PEf

1575.31 J +0 =0 + mgh

Solving for h , we get

h = 2.57 m

(e) At half height point person has both KE and PE .

Solving

KEi + PEi = KEf + PEf

(1/2)mu2 +0= mgh + (1/2)mv2

putting u = 7.1 m/s ; v = 5.02 m/s

We get h = 1.285 m

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