Question

A box of relief good is dropped from rest from a helicopter hovering 25.0m above the ground. What is the speed of the box just before it reaches the ground?

Answer 1

Explanation:

speed= d/t

d is given

to find t

s=ut +1/2 gtsquare

s is given

g is 10mper second square

u is 0

find out t

then put the value

Answer 2

Answer:

22.36 m / sec

Explanation:

Given :

Initial velocity u = 0 m / sec

Final velocity v = ? m / sec

Distance h = 25 m / sec

From third equation of motion :

v² = u² + 2 g h

v² = 0 + 2 × 10 × 25

v² = 500

v = √ 500

v = 22.36 m / sec

Therefore , speed of the box should be 22.36 m / sec just before it reaches the ground.