physics problem can nyone help?

Question

If a 5kg mass falls a 150 m tall building, how fast it going the instant it hits the ground?

can u also show me how to do it??? not juz da answer

Answer 1

the acceleration of gravity is 9.8m/s/s. This is a constant for the earth. Many physics texts use 10m/s/s to make the calculations easier.

What this means is that as the object falls, it's speed increases by 10m/s every second. So after one second, speed is 10m/s, after two seconds, it's 20m/s, etc.

We can use the followng formula to calculate the answer:

d= v(i)t+1/2(at^^2)

where d=distance, in this case 150M
v(i)= the initial velocity of the object, in this case, zero
a= acceleration, 10m/s/s
t= time

Plugging in our numbers, and solving:

150= 0T+ 1/2(10T^^2)
150=5T^^2
30=T^^2
sqrt(30)=T
5.47=T

Now we know that it takes 5.47 seconds to hit the ground. Since acceleration is 10m/s/s, final velocity is 54.7m/s.

You may have noticed that the mass wasn't taken in to consideration. This is not by accident: for anything small relative to the whole earth, the acceleration of gravity is a constant: 10m/s/s. So whether the object weighs 5kg or 1000kg, the acceleration will still be 10m/s/s.

Answer 2

Easy.

9.81 m/s/s = g is the acceleration of gravity.

That is 9.81 m/s/s or 9.81 meters per second per second. Interestingly, first of all, the mass of the object DOES NOT MATTER. I will assume you mean a sphere to reduce drag or that there is no atmosphere as that calculation is complicated, but if you need to take wind resistance into account, I suggest using a sphere.

So, the equation you need is to compute the velocity of an object accelerated by 9.81 m/s/s over a 150m distance. Acceleration is given by the change in velocity over the change in time, a.k.a. dv/dt. (t is time, v is velocity). Since you know the acceleration (it is gravity at 9.81 m/s/s), if you know the time traveled, you can compute the velocity via dv = 9.81 m/s/s * dt.

Now, we know that velocity (speed) is the change in distance over the change in time: dl/dt. Here, we know the distance, 150 m, but we can not plug that value in directly. This is because although we know the time it takes to travel a given distance at a given velocity, the velocity is not constant so we can't use this result directly.

What we first need to do is calculate the travel time. We do this by combining our two expressions:

v = dl/dt
dv = 9.81 m/s/s * dt

Combining:
ddl/dt = 9.81 m/s/s * dt

(Note the two d's, one from dl, the other from the d in front of dv).

Here, we again multiply by dt:

ddl = 9.81 m/s/s * dt * dt

Finally, we apply something called integration. This is symbolized by a tall, narrow letter s, as was used to write the letter s 200 years ago (see the U.S. Declaration of Independence for an example). Here I will use | for the integral symbol:

| ddl = 9.81 m/s/s | dt*dt
| ddl = | dl = l
| dt*dt = | t dt = 1/2 t*t

Thus,

l = 150m = 1/2 * 9.81 m/s/s * t*t,

or:

t = sqrt(150m * 2 / (9.81 m/s/s))
t = sqrt(30.6) s
t ~= 5.53 seconds

Where sqrt is the square root operation.

We can now solve for the velocity:

dv = 9.81 m/s/s * dt
dv ~= 9.81 m/s/s * 5.53 s
dv ~= 54.2 m/s

So your speed when your 5kg object hits the ground is approximately 54.2 meters per second or 118 feet per second or 121 miles per hour or 195 km/h.

Hope that helps!

Answer 3

The mass is irrelevant.
One usually ignores air friction in such problems because the terminal velocity is solely dependent on the shape of the object.
The formula is: s = 1/2 at^2
s = distance
a = acceleration (9.8 m/sec^2)
t = time
150 = 1/2 * 9.8 * t ^ 2
t = sqrt( 300 / 9.8)
t = 5.5 sec
velocity would be 9.8 * 5.5 = 53.9 m/sec

Answer 4

The previous answer is correct except for the negative sign for acceleration due to gravity.

In the problem all the velocities distance and accleration due to gravity are taken to be positive when falling.

Answer 5

Well if it starts from rest then...

Vf^2 = Vi^2 + 2ax

Vf^2 = (0^2) + (2)(-9.81)(150)

Vf = sqrt (2943)

Vf = 54.2 m/s

the -9.81 is the acceleration due to gravity