Question

A child throws a ball vertically upward to a friend on a balcony 28 m above him. The friend misses the ball on its upward flight but catches it as it is falling back to earth. If the friend catches the ball 3.0 s after it is thrown, at what time did it pass him on its upward flight

Answer 1

Answer:

$t=1.9 sec$

Explanation:

From the question we are told that:

Height $h=28m$

Time $t=3s$

Generally the Newton's equation for Initial velocity upward is mathematically given by

 $s=ut+\frtac{1}{2}at^2$

 $28=3u-\frac{1}{2}*9.8*3^2$

 $u=24.03m/s$

Generally the velocity at  elevation and depression occurs  as ball arrives and passes through S=28

 $v=\sqrt{24.03-2*9.8*28}$

 $v=5.35m/s and -5.35m/s$

Generally the Newton's equation for time to reach initial velocity  is mathematically given by

 $v=u+at$

 $5.35=24.03-9.8t$

 $t=\frac{28.03-5.35}{9.8}$

 $t=1.9 sec$