#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# A ball is thrown from a point with a speed V  at an angle of projection  θ . From the same point and  at the same instant, a person starts running with a constant speed V/2 to catch the ball. Will the  person be able to catch the ball? If yes, what should be the angle of projection?

upender surepally
126 Points
8 years ago

Man will catch the ball if the horizontal component of velocity becomes equal to the constant speed
of man i.e. θ=600

Dhiraj Bhakta
18 Points
8 years ago

horizontal distance covered by the ball= horizontal range= (hope u know the formula) R=[V2sin2(theta)]/g

time taken by the ball to reach the ground=T=2Vsin(theta)/g

note;] remember that time taken by the person to reach the spot and time taken

by the ball to reach the ground(same spot)..is THE SAME;

time taken by the person to reach the spot= time taken by the ball..

distance/speed (man)          =        2v sin theta/g  [ball]

range/speed[man]               =             '''' ''''

( V2sin 2 theta/g) / v/2        =              '''' ''''

things cancel out in the equation and finally u get     sin (theta) = sin 2 (theta)

this can happen if theta = 00 ...............................which is not possible

...therefore  2 theta= 180-theta.........by trigonometry

3 theta = 180           =====>theta=600

Yashdeep singh
15 Points
2 years ago
A ball is thrown from a point with a speed V  at an angle of projection  θ . From the same point and  at the same instant, a person starts running with a constant speed V/2 to catch the ball. Will the  person be able to catch the ball? If yes, what should be the angle of projection?
riya
13 Points
2 years ago
Yes, the person will be able to catch it at theta=60 degree.horizontal component of velocity becomes equal to speed.
11 months ago
Dear student,

For the boy to catch the ball, he must travel the range of projectile(ball) in the same time as the time of projectile.
Now, there is no horizontal direction acceleration.
Hence, vcosθ x T = v/2 x T
or, cosθ = ½
or, θ = 60o
Hence he will be able to catch the ball only when the angle of projection is 60o

Thanks and regards,
Kushagra