Get Answers to all your Questions

header-bg qa

Q. 4.15 The ceiling of a long hall is 25\; m high. What is the maximum horizontal distance that a ball thrown with a speed of 40\; m \; s^{-1} can go without hitting the ceiling of the hall?

Answers (1)

best_answer

It is known that the maximum height reached by a particle in projectile motion is given by  : 

                                                   h\ =\ \frac{u^2sin^2\Theta }{2g}

Putting the given values in the above equation :

                                                  25\ =\ \frac{40^2sin^2\Theta }{2\times9.8}

So, we get                               

 \sin \Theta \ =\ 0.5534      and           \Theta \ =\ 33.60^{\circ}

Now the horizontal range can be found from   : 

                                                 R\ =\ \frac{U^2 \sin 2\Theta }{g}

  or                                                   =\ \frac{40^2 \sin (2\times 33.6 )}{9.8}

                                                         =\ 150.53\ m 

Posted by

Devendra Khairwa

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads