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The formula is t = 2u/g. U is 49m/s. It is independent of velocity of the lift. the acceleration is g only. Hence the answer is 10 s

The formula is t = 2u/g. U is 49m/s. It is independent of velocity of the lift. the acceleration is g only. Hence the answer is 10 s

Body projected vertically up at the highest point possess zero velocity, but it experiences acceleration namely acc. due to gravity. Pl rate if satisfied.

not possible. Because every physical quantity depends on the same fundamental quantities which are raised to same powers.

aman,
There is nothing like absolute rest or absolute motion. Both are relative terms. They are expressed with respect to some reference surroundings. In fact all the terms are relative only like good or bad, hard or soft, hot or cold etc.

focus on Terminal velocity:

Are the rain drops harmful?

   Yes, had there not been atmosphere around the earth, the rain fall must have been very dangerous for us. The atmosphere around the earth is like helmet. See the simple example to show this.

          Assume the rain to be falling from the clouds from a height, 1000 m i.e 1 K.M. Velocity on reaching the ground will be v2.g.h = v2 . 9.8 . 1000 = 140 m/s = i.e nearly 450 K.M.P.H i.e the speed of a bullet train. If each drop strikes us with this velocity probably every body must have died long back. This is not happening because the air molecules around us produce retardation for the rain drops and make them attain bearable low velocity called terminal velocity. Parachutist also lands safely without hurting himself, based on this principle. If air molecules are absent, it makes no difference whether he wears a parachute or not. Is it not .

          Note: The value of g is assumed to be constant throughout the distance of 1 km.

          Yr comments on this article are welcome.

u r saying nothing to do with friction, and u r saying coefficient of friction reduces. When friction is not there where is the qn of coefficient of friction. pl check ur answer again. leave it for others to react. cheers

r these stuff down loaded from internet sites? If so pl put the things with ur own effordt

The equation for time of descent is t =

2h/g. Therefore, t₁ /t₂ =

a/b

Use the general equation u = u_x i + u_y j

ucos

= 6 m/s and u sin

= 8 m/s
Range =2 ucos

u sin

/g Hence the range is 9.6 m

prakar, do u really think that two bodies do not come in contact at all? How wheels of vehicle roll on the road, not being in contact with the road? Pl check

When the body is pulled up the plane, the friction and the component of weight mg sin

act down the plane. Since the motion is constant, same upward force must act up the plane given by f = 2000 + mg sin

= 51000 N. The power is given by

P = f * V = 510000 W = 510 KW. figure i have drawn on paint brush. I am not able to paste it here. I can mail it if u give ur mail adress

Yes initial velocity is + 2m/s which indicate upward motion. Naturally the final velocity will be negative. Which shows its direction finally is - Ve. to find final velocity, when it hits the ground, use the equation V² - U² = 2 (-g) (-h). Here the displacement downwards i.e 24 m is negative as it is measured from top to bottom. ( ground ). Pl rate this explanation.

check the answer

t = 320 / Vm

Vm = 320 / 4 = 80

Vm = 80

Vm = 5 Vr/3

Vr = 80 * 3/ 5 = 48 m/ min. If in the problem it is mentioned as 5/3 with respect to water, the procedure given by nishant is correct

Power is given by P = W/ t
                            = (force * disp)/t   Multiply and divide with A (area)
                              = (f * s) A/A
                            = p * V/t    Because, f/A is pressure, p and s * A is volume V
                              = pressure * volume/time. Here p = hdg = 0.13 * 4 * 10³ 9.8
To get power in H.P divide with 746.