Answer to Question #170488 in Physics for Prince

Question

A ball thrown vertically upward reaches a maximum height of 50m above the level of projection. Calculate the

1. time taken to reach maximum height
2. Speed of the throw

Answer 1

The speed of the throw can be found from the energy conservation law: the initial kinetic energy of the ball is equal to its potential energy.

$$\dfrac{mv^2}{2} = mgh\\$$

where $m$ is the mass of the ball, $v$ is its initial speed, $h = 50m$ is the maximum height, $g = 9.8N/kg$ is the gravitational acceleration. Expressing the speed, obtain:

$$v = \sqrt{2gh}\\ v = \sqrt{2\cdot 9.8\cdot 50} \approx 31.3m/s$$

The distance covered by the ball is given by the kinematic equation:

$$h = vt - \dfrac{gt^2}{2}\\ \dfrac{gt^2}{2} - \sqrt{2gh}t+h=0$$

where $t$ is the time.

Solving the quadratic equation, obtain:

$$D = 2gh-2gh = 0\\ t = \dfrac{\sqrt{2gh}}{g} = \sqrt{\dfrac{2h}{g}}$$

When $h = 50m$, the time will be:

$$t = \sqrt{\dfrac{2\cdot 50}{9.8}} \approx 5.1s$$

Answer. 1) 5.1s, 2) 31.3 m/s.