A cricket ball of mass 0.15\; kg is thrown vertically up, by a bowling machine, so that it rises to a maximum height of 20\; cm after leaving the machine. If the part pushing the ball applies a constant force F on the ball and moves horizontally a distance of  0.2\; m while launching the ball the value of F (is N). is (g=10\; m/s^{2})
Option: 1 10
Option: 2 100
Option: 3 150
Option: 4 400

Answers (2)

Work done by bowling machine = max. P.E. of the ball

\begin{array}{l} F \times s=m g h \\ F=\frac{m g h}{s}=\frac{0.15 \times 10 \times 20}{0.2}=150 N \end{array}

 

Most Viewed Questions

Preparation Products

Knockout JEE Main (Six Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 9999/- ₹ 8499/-
Buy Now
Knockout JEE Main (Nine Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 13999/- ₹ 12499/-
Buy Now
Test Series JEE Main 2024

Chapter/Subject/Full Mock Tests for JEE Main, Personalized Performance Report, Weakness Sheet, Complete Answer Key,.

₹ 7999/- ₹ 4999/-
Buy Now
Knockout JEE Main (One Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 7999/- ₹ 4999/-
Buy Now
Knockout JEE Main (Twelve Months Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 19999/- ₹ 14499/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions