Thank you for registering.

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

Please check your email for login details.
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 projectile of mass 2m is projected at an angle of 45 degrees with the horizontal with a velocity 20root[2]m/s. After 1 s explosion takes place and the particle is broken into two equal pieces. As a result of explosion one part comes to rest. Find the maximum height attained by the other part. Take g=10m/s^2.

A projectile of mass 2m is projected at an angle of 45 degrees with the horizontal with a velocity 20root[2]m/s. After 1 s  explosion takes place and the particle is broken into two equal pieces. As a result of explosion one part comes to rest. Find the maximum height attained by the other part. Take g=10m/s^2.

Grade:11

1 Answers

Swapnil Saxena
102 Points
9 years ago

The answer for this question is 35 m.

Initially the velocity is 20root(2)ms-1 & the angle is 45. Vertical component of velocity = 20ms-1

After 1 s, It will cover a distance of s = ut+1/2at2 = (20)(1)+1/2(-10)(1)2 = 20 - 5 = 15m and its vertical velocity comp. will be reduced to 10ms-1.

Soon After it s mass gets reduced to m from 2m due to reaking so according to the law of conser. of center of mass,  it velocitys vertical comp. will be doubled. = 10ms-1.*2 = 20ms-1

Now for coming to rest it has to cover diatnce v2= u2-2gs=> 400/20=s = 20m

So the total height achieved by the the ball is 20+15 = 35 m


Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free