PHYSICS Q help?

Question

1} A body is dropped from a baloon which is moving upwards with aspeed of 10m/s.the body is drooed when the baloon is at height =390m from earth surface.find time for the body to reach the surface of the earth??
PLEASE GIVE ME A DETAILED ANS!!

2] what will area under dis-time graph give?

g=10m/s^2
please help me

Answer 1

Using the relationship v^2 - u^2 = 2as, we can compute the distance s moved upwards by the body due to the initial velocity not being zero. This distance and the time taken are to be added to the original height of the balloon from the earth when the body was released and the time taken from there to fall down.

u = 10 m/s a = -10 m/s^2 till the body reaches 0 velocity. Let s be the distance it moves.

0 - 10 X 10 = 2 X -10 X s or s = 100 / 20 = 5 metres.

Let t1 be the time taken for this upward motion (at the end of that, it comes to a momentary rest and then starts dropping down).

s = ut1 + 1/2 at1^2

5 = 10.t1 + 1/2 -10.t1^2 = 10t1 - 5t1^2 (a = -g = -10m/sec^2)

or 1 = 2t1 -t1^2 (dividing by 5) and this can be written as

t1^2 -2t1 + 1 = 0 or (t1 - 1)^2 = 0 and t1 = 1 sec.

Add the 5m to the original height 390 giving a height of 395m. From there, it is a free fall to earth. Let t2 be the time for the body to reach earth.

Again we use the formula s = u.t + 1/2.a.t^2

395 = 0.t22 + 1/2.10.t2^2 (. indicates multiplier)

395 = 5t2^2 or t2^2 = 395/5 = 79

t2 = sqrt.79 = 8.8882 secs.

Total time t = t1 + t2 = 1 + 8.8882 = 9.8882

So total time is 9.8882 secs.

2) A look at the distance - time (distance is on y axis and time is on x axis) will tell us whether the motion is at a constant velocity or with varying velocity. In this case, we have initially an upward motion of 5 metres and then a downward motion of 395 metres. So, the distance will be initially negative (if we take the upward motion as negative) and then will cross the x-axis and will go upward with uniform acceleration. If we take the distance from the ground as positive, the graph will start with y = 390 at t = 0, move up to 395 when t = 1, and then start the downward curve and hit the x-axis (y = 0) at t = 9.8882 secs.

Answer 2

Nice question. I'll assume this is a simple question that does NOT involve the calculation of frictional drag and trajectory (The angle at which the body was ejected)

1) If the body was simply dropped from a balloon that was travelling vertically upwards at 10 m/s, then the first part is akin to launching a body at an initial velocity of 10 m/s from the ground vertically upwards.

V=velocity at time t, u=initial velocity and g=gravity (9.8m/s^2)

Using V=ut - gt first and then S=ut-1/2gt^2

0=10t-9.8t, => t=1.02s S=10t-4.9t^2, => S=5.1m


Therefore, the body will travel upwards for 1.02 seconds up to a height of 5.10m+390m=395.1m before falling freefall.

For the second part, freefall time calculation, use

S=1/2 gt^2 using S=395.1m => 8.98s

So, the whole time spent from dropping the body to reaching the ground = 1.02+8.98= 10s

2) There is no stated initial departure point of reference from which to make a distance-time graph. Should it be from the point the body was dropped (390m above the ground), or should it be from the point the balloon took off from the ground? If it is the former, then it will start at t=0, d=0, to t=1, d=-5.01 and from that point to t=10, d=390. Join up these coordinates and work out the area under the graph with simple trigonometry.

=> The area in the -ve y axis (When the body was travelling upwards then going down again to the initial point of launch) = 0.5x2.04x5.1=5.2

And the area in the +ve y axis when the body was travelling in freefall from the initial point of launch to the ground = 0.5x7.96x390=1552.2

Answer 3

The area under the graph can be described by the definite integral of the function.

Answer 4

s=ut+0.5at^2
s=390
u=-10
g=+10


so t=9.88 s
on solving the equation with substituin the values


integtat of S*dt from 0 to 9.88
gives the area as 1119.311

Answer 5

You kinda spoke back your guy or woman question. it is experiential and relative to the guy. The longer we are residing, the greater genuinely we start to understand longer and longer instruments of time as a results of fact we've experienced them. as a effect, the 'shorter' instruments of time become something that 'look' to bypass greater right away as our unsleeping concentration turns into greater attuned to longer periods. Doug