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?
Answers
Answered by
0
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
Answered by
5
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.
Similar questions