In the high jump, the kinetic energy of an athlete is transformed into gravitati

Question

In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole. With what minimum speed must athlete leave the ground in order to lift his center of mass 1.90 m and cross the bar with a speed of 0.95 m/s? In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole. With what minimum speed must athlete leave the ground in order to lift his center of mass 1.90 m and cross the bar with a speed of 0.95 m/s? With what minimum speed must athlete leave the ground in order to lift his center of mass 1.90 m and cross the bar with a speed of 0.95 m/s?

Explanation / Answer

Kinetic Energy + Potential Energy increase

0.5 mv^2 + mgh

g is gravity

h is height

v is velocity

so

TE = 0.5 *m*(0.9)^2 *+ m*g*1.90

Now while leaving the ground,

Kinetic Energy = 0.5mv^2

Therefore, multiplying by 2/m,

v^2 = 0.9^2 + 3.8 g

V = 6.16 m/s

Not much more than the 6.16 m/s required merely to lift 1.9m and have zero speed at the top

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.