Question

An athlete on a trampoline leaps straight up into the air with an initial speed of 7.0 m/s. Find the following: (a) the maximum height reached by the athlete relative to the trampoline m (b) the speed of the athlete when she is halfway up to her maximum height m/s

Answer 1

The maximum height (h) reached by the athlete is 2.50 m. At halfway up to the maximum height, the athlete's speed is approximately 5.48 m/s.

Step-by-step explanation:

The problem presented is related to the physics of motion, specifically projectile and free-fall motion. To solve part (a), we use the kinematic equation that relates initial velocity, acceleration due to gravity, and maximum height:
h = (v^2) / (2g), where h is the maximum height, v is the initial speed, and g is the acceleration due to gravity (9.81 m/s2 on Earth).

Substituting the given values we get: h = (7.0 m/s)2 / (2 × 9.81 m/s2) = 2.50 m.

For part (b), when the athlete is halfway up, the vertical speed can be found using energy conservation principles. The kinetic energy at any point is equal to the initial kinetic energy minus the potential energy gained. Therefore, at half height, 1/2 m(v_half)^2 = 1/2 m(v_initial)^2 - mgh/2.

Solving for v_half, we find v_half = √(v_initial^2 - gh). Plugging in the values, v_half = √((7.0 m/s)^2 - 9.81 m/s2 × 1.25 m) ≈ 5.48 m/s.