A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity.

Answers (1)

Initial velocity u = 0

Acceleration, a = g = 9.8 m s-2

Distance travelled, s = 19.6 m

Let the final velocity be v

According to the third equation of motion

\v^2-u^2=2as\ v^2-0^2=2	imes 9.8	imes 19.6\ v=sqrt2	imes 9.8	imes 19.6\ v=19.6 m s^-1

Its final velocity just before touching the ground will be 19.6 m s-1.

 

Most Viewed Questions

Preparation Products

Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 40000/-
Buy Now
Knockout NEET 2025

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 45000/-
Buy Now
NEET Foundation + Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 54999/- ₹ 42499/-
Buy Now
NEET Foundation + Knockout NEET 2024 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
NEET Foundation + Knockout NEET 2025 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions