Answer to Question #202160 in Optics for Ibrahim

Question

A ball is dropped from rest out of a high window in a tall building for 5 seconds. Assuming we ignore air resistance and assume upwards to be positive. A) What will be the final velocity of the ball B) What is the height of the building if it hits the ground after those 5 seconds.

Answer 1

A.) Given,

$Initial\ velocity\ u=0\\Final\ velocity=v\\g=-9.8\ m/s^2\\time\ t=5\ seconds$

So,

$v=u+gt\\\implies v= 0+(-9.8)(5)\\\implies v=-47.5\ m/s$

Hence, final velocity of ball is 47.5 m/s downward.

B.) Height of building

$h=ut+\dfrac{1}{2}gt^2\\\because u=0\ and\ t=5$

$h=0+\dfrac{1}{2}(-9.8)(25)\\\implies h=-122.5\ m$

Hence, the height of building is 122.5 m