Question

a ball is dropped from rest at a height of 45.0 m above the ground. ignore the effects of air resistance. What is the speed of class of the ball when it is 10.0 m above the ground

Answer 1

So, the final velocity of the ball when it is 10.0 m above the ground approximately 26.2 m/s.

Introduction

Hi ! In this question, I will help you. This question uses the principle of final velocity in free fall. Free fall occurs only when an object is dropped (without initial velocity), so the falling object is only affected by the presence of gravity. In general, the final velocity in free fall can be expressed by this equation :

$\boxed{\sf{\bold{v = \sqrt{2 \times g \times h}}}}$

With the following condition :

• v = final velocity (m/s)
• h = height or any other displacement at vertical line (m)
• g = acceleration of the gravity (m/s²)

Problem Solving

We know that :

• $\sf{h_1}$ = initial height = 45.0 m
• $\sf{h_2}$ = final height = 10.0 m
• g = acceleration of the gravity = 9.8 m/s²

Note :

At this point 10 m above the ground, the object can still complete its movement up to exactly 0 m above the ground.

What was asked :

• v = final velocity = ... m/s

Step by Step

$\sf{v = \sqrt{2 \times g \times \Delta h}}$

$\sf{v = \sqrt{2 \times g \times (h_1 - h_2)}}$

$\sf{v = \sqrt{2 \times 9.8 \times (45 - 10)}}$

$\sf{v = \sqrt{19.6 \times 35}}$

$\sf{v = \sqrt{686}}$

$\boxed{\sf{v \approx 26.2 \: m/s}}$

Conclusion

So, the final velocity of the ball when it is 10.0 m above the ground approximately 26.2 m/s.

See More :

• The relationship between acceleration and the change in velocity and time in free fall brainly.com/question/26486625