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 particle is thrown vertically upwards.Its velocity at one fourth of maximum height is 20m/s.Then the max height is

A particle is thrown  vertically upwards.Its velocity at one fourth of maximum height is 20m/s.Then the max height is

Grade:10

2 Answers

vikas askiitian expert
509 Points
10 years ago

let H is maximum height ...

if initial velocity is u then

applying , v2 = u2 + 2as

at maximum height , v=0 , s=H , a=-g   ,after putting these

         0 = u2 -2gH     or  u2 = 2gH    ...................1

again applying , v2 = u2 + 2as upto that instant when its velocity was 20m/s...

   s = H/4 , v = 20m/s

           202 = u2 - 2g(H/4)         .........2

from 1 putting u2 in 2

        202 = 2gH - 2g(H/4)

       400 = 2gH/3

      H=80/3 = 26.7m

approve if u like my ans

Fawz Naim
37 Points
10 years ago

if the maximum height is H then at one-fourth of this height

v^2=u^2-2gH/4, v=20m/s

400+2gH/4=u^2 .....1

now at the heighest point v=0

v^2=u^2-2gH

0=u^2-2gH

u^2=2gH

putting value of u^2 from equation 1

400+2gH/4=2gH

H=80/3 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