a particle is thrown vertically upwards..............................?

Question

a particle is thrown vertically upwards.if its velocity at half the max.height is 10m/s,then max.height attained by it is?(g=10m/s^2)

Answer 1

Well it will take 1 second to stop the final bit of velocity:

S=ut+0.5at^2

S=10+(0.5*-10*1^2)

S=5m

Total height is double this as you said it was halfway up, 10m.

Answer 2

First you're able to locate out the cost of the 1st particle after 2 seconds. utilising the equation init. veloc= very final veloc + (accel)(time) and plugging in 20 m/s for the preliminary, 10 m/s^2 for acceleration and a couple of.0 seconds for time, you come across that when 2.0 seconds, the cost of the 1st particle is 0 for it has reached the suitable of its path. next you're able to set equations for the placement of each and every particle equivalent to one yet another. Use the equation: place= (init. veloc)(time) + (one million/2)(accel.)(time^2). Your new preliminary is after 2 seconds. Set those place equations equivalent to one yet another, utilising 0 m/s for the preliminary velocity of the 1st particle and 20 m/s for the different. determine you reside consistent with the sign you employ on acceleration and velocity. (as an occasion in case you make certain that an merchandise vacationing up is going interior the detrimental path, this is velocity is detrimental. locate the sign for acceleration to that end.) then you certainly in simple terms resolve for time. wish this helps!

Answer 3

v^2 - v0^2 = 2as
v at max height = 0
2(10)(s/2) = 10^2
s = 100/10 = 10 m

Answer 4

v=-10t +vo
h=-5t^2+vot
The maximum height occurs when -10t +vo=0 so t =vo/10 and
hmax=-5(vo^2)/100+vo^2/10 = vo^2/20
1/2hmax = vo^2/40 and occurs when

10=-10t+vo so t= (vo-10)/10 and putting this time in h
vo^2/40=-(vo-10)^2/20+vo(vo-10)/10
vo^2/4=(vo-10)(1/2+vo)
This is a 2nd degree equation for vo and vo^2/20 gives you the maximum height (vo must be >0)