Question

A body projected Vertically upwards with a velocity v at t=0 is found at a height h after 1 second and after a further 6 second it is found at the same height (g=10m/s^2) then

h=30m

u=40

Hmax=80m

distance moved in 5th second is 5m

this is a multiple answer question more than one answer is correct.

Answer 1

vat is u here?

Answer 2

Not given?

Answer 3

i thnk it must b v.

Answer 4

all are correct

Answer 5

@Arnab09 no answer is b c d

Answer 6

okay, lemme check again

Answer 7

if b is true, a is ought to be true

Answer 8

No......@Arnab09

Answer 9

lemme check again, h=ut- 1/2 gt^2.. right?

Answer 10

what a fool i am.. h has to be 35 m

Answer 11

b,c,d
why r u posting these questions, @open_study1 ?

Answer 12

I wanted u knw the steps to do this

Answer 13

apply basic formulae, thats all

Answer 14

only trick is here: after 6 seconds it returns to the same height.
so, it will take 6/2=3 seconds more to reach the Hmax..
so, it will take 4 seconds to reach Hmax
at Hmax, v=0
so, u can apply 0=u-gt
find u and u can do the rest, i believe

Answer 15

@Arnab09 i will try my best to solve this plzz wait and will inform u..

Answer 16

i got Hmax=80m

Answer 17

and u as 40m

Answer 18

Distance in 5 second ???

Answer 19

5th second

Answer 20

u+ 1/2 a(2t-1)
u will find it by calculation