Question

A ball is thrown straight upward. At 4.60 m above its launch point, the ball’s speed is one-half its launch speed. What maximum height above its launch point does the ball attain?

PLEASE POST WITH EXPLANATION. Will give points to well-explained post with steps.

Answer 1

To find the maximum height above its launch point that the ball attains, we need to use the kinematic equation for vertical motion.

Step-by-step explanation:

To find the maximum height above its launch point that the ball attains, we need to use the kinematic equation for vertical motion.

The equation is: v^2 = u^2 - 2gs

Where v is the final velocity (which is 0 when the ball reaches the maximum height), u is the initial velocity, g is the acceleration due to gravity, and s is the displacement.

Since the ball reaches a height of 4.60 m above its launch point, the displacement is 4.60 m. The acceleration due to gravity is approximately 9.8 m/s^2.

Plugging in these values, we get: 0 = u^2 - 2(9.8)(4.60)

Simplifying the equation, we have: u^2 = 2(9.8)(4.60)

Calculating, we find that u ≈ 9.62 m/s, which is the initial velocity of the ball.

Answer 2

The maximum height above the launch point the ball attains can be found using conservation of energy principles. Since the ball's speed is halved at 4.60 m, implying one-fourth of its initial kinetic energy, the maximum height is four times this potential energy height, resulting in 18.4 m above the launch point.

Step-by-step explanation:

The subject of this question is Physics, specifically it involves concepts from kinematics and projectile motion. To determine the maximum height above its launch point that the ball attains, we can utilize the principles of conservation of energy. If the ball's speed at 4.60 m is half of its initial launch speed, the loss in kinetic energy must have been converted into potential energy for the ball to rise to that height. Without the need for actual numbers, we can deduce that at the maximum height, the ball has zero velocity and all its initial kinetic energy has been converted to potential energy.

To preserve the energy, the relationship (1/2)mv_i² = mgh is used, where v_i is the initial launch speed, g is the acceleration due to gravity (approximately 9.81 m/s²), and h is the maximum height. Because the speed at 4.60 m is half of the initial speed, we can write (1/4)mv_i² = mg(4.60 m), indicating that the potential energy at 4.60 m is a quarter of the initial kinetic energy.

This means that the ball must have enough initial kinetic energy to reach four times that potential energy height, so the maximum height (h) will be 4 * 4.60 m = 18.4 m above the launch point, assuming that all initial kinetic energy converts to potential energy at the peak.