Question

Alex climbs to the top of a tall tree while his friend Gary waits on the ground below. Alex throws down a ball at 8 m/s from 50 m above the ground at the same time Gary throws a ball up. At what speed must Gary throw a ball up in order for the two balls to cross paths 25 m above the ground? The starting height of the ball thrown upward is 1.5 m above the ground. Ignore the effects of air resistance. whats the answer in m/s?

Answer 1

Answer:22.62 m/s

Explanation:

Given

two balls are separated by a distance of 50 m

Alex throws  the ball from a height of 50 m with a velocity of 8 m/s and Gary launches a ball with some velocity  exactly at the same time.

ball  from ground travels a distance of 25 m in t sec

For Person on tree  

$25=ut+\frac{1}{2}gt^2$

$25=8t+\frac{1}{2}\times 9.81\times t^2--------1$

For person at ground

$23.5=ut-\frac{1}{2}gt^2---------2$

Solve equation (1)

$50=16t+9.81t^2$

$9.81t^2+16t-50=0$

$t=\frac{-16\pm\sqrt{256+4\times 50\times 9.81}}{2\times 9.81}=\frac{47.1-16}{19.62}=1.58 s$

put the value of t in equation 2

$23.5=u\times 1.58-\frac{9.81\times 1.58^2}{2}$

$u=\frac{35.744}{1.58}=22.62 m/s$