Question

a particle is thrown upwards.it's velocity at half of the height is 10m/s,then calculate the maximum hieght attained by it

Answer 1

We know acceleration = -9.8 m/s^2 and final velocity = 0 m/s. If we treat 10 m/s as the initial velocity, the total distance it travels until its velocity reaches 0 is HALF the maximum height, since it is traveling at 10 m/s HALFWAY through its trip.

Use vf^2 = vi^2 + 2ad:

0 = 10^2 - 2(9.8)(0.5hmax)
hmax = 10.2 m

Answer 2

my answer is also same as above one, i..e., max. height attained= 10.2m

Answer 3

listen.... i am a 9th grade and cannot understand your equations can someone give me a answer which i can understand ....plz.......i have to submit my homework....

Answer 4

do you know the three equations if motion for constant acceleration?
i.e. v = u + at
s = ut + 0.5at^2
v^2 - u^2 = 2as

Answer 5

no.....i don't just explain plz.

Answer 6

whats s,u,a,t are they short forms

Answer 7

u = intial velocity,
v= final velocity
a = acceleration
t=time
s = displacement

Answer 8

and what do you mean by this thing ^2

Answer 9

its the exponent .. x^y = x raised to y

Answer 10

so it means square of something

Answer 11

Do you understand how we got the answer? I'll rewrite my answer using the equation tool to make it clearer if you need and provide more explanation.