Question

if a ball is dropped from a height 20 metre with valency 5 metre per second at same another ball is thrown from ground at a speed of 20 metre per second find its distance from ground when they meet​

Answer 1

Answer:

0.8 Seconds

Let the height be x .

Use

$s = ut + \frac{1}{2} a {t}^{2}$

For ball B

s = x , u = 20 , a = -10

$x = 20t - \frac{1}{2} \times 10 {t}^{2} \\ x = 20t - 5 {t}^{2}$

For ball A

At t = t , distance travelled is 20 - x .

u = 5 , g = +10

$20 - x = 5t + \frac{1}{2} g {t}^{2}$

Substitute value of x

$20 - (20t - 5 {t}^{2} ) = 5t + 5 {t}^{2} \\ \\ 20 - 20t + 5 {t}^{2} = 5t + 5 {t}^{2} \\ \\ 25t = 20 \\ \\ t = \frac{20}{25} = \frac{4}{5} \: s = 0.8 \: s$