Answer to Question #170059 in Mechanics | Relativity for Stephen Ogu

Question

A ball is projected vertically upward with a velocity of 80m/s. Two seconds later a second ball is projected vertically upward with a velocity of 60m/s. At what point above the surface of earth will they meet?

Answer 1

Let's take the upwards as the positive direction. After $t=2\ s$ the first ball reached the height $y(2)$ and its vertical displacement can be written as follows:

$y_1=y(2)+v(2)t-\dfrac{1}{2}gt^2.$

Let's find the velocity and height of the first ball after $t=2\ s$:

$v=v_0-gt=80\ \dfrac{m}{s}-9.8\ \dfrac{m}{s^2}\cdot2\ s=60.4\ \dfrac{m}{s},$ $y(t=2\ s)=80\ \dfrac{m}{s}\cdot2\ s-\dfrac{1}{2}\cdot9.8\ \dfrac{m}{s^2}\cdot(2\ s)^2=140.4\ m.$

Then, we can write the vertical displacement of the second ball projected vertically upward two seconds later:

$y_2=v_{02}t-\dfrac{1}{2}gt^2$

At the moment when they meet $y_1=y_2$ and we can write:

$y(2)+v(2)t-\dfrac{1}{2}gt^2=v_{02}t-\dfrac{1}{2}gt^2.$

Substituting $y(2)$ and $v(2)$ in the previous equation, we get:

$140.4+60.4t-\dfrac{1}{2}gt^2=60t-\dfrac{1}{2}gt^2,$ $140.4=-0.4t$ $t=-351\ s.$

The time is negative, which means that two balls never meet at the height above the earth.