Elevator problem. help please

[jhson114]

the problem looks pretty straight forward, but i suck at physics and i cant solve it :(

a person is inside an elevator going up at constant speed of 5.7m/s. person drops a ball at height 1.8m. how long does it take the ball to hit the elevator floor?

i set up my equation thingy to 0=1.8 +5.7t + 0.5(-9.8)t^2 but i think its wrong. help please

[Chi Meson]

The hint here is "Constant speed." ANd since the elevator travels in a srtaight line, this also implies "constant velocity."

Whenever something is traveling at a constant velocity, it is said to be in an "inertial reference frame," And anything that happens in this frame will act just as if you wer standing on the ground.

So, what the ball does in the elevator will be identical to what a ball does when you are on the ground.

[jhson114]

i see. so the speed of elevator isnt important in this situation?

[jhson114]

i got it. haha. tricky tricky. thanks :)

[Chi Meson]

No problem.

[Naeem]

I didn't get this problem,

I tried doing :

x -x0 = vot + 1/2 at^2

constant speed means acc = 0

so, h = 2.1 m and constant speed v0 = 4.6 m/s ( In my case )

Time = 2.1/4.6

= 0.45 sec. which is wrong??? Any help

[codyg1985]

The only thing to consider for this problem is the acceleration due to gravity and the height of the ball from the bottom of the elevator.

The speed of the elevator is irrelevant because its acceleration is zero.

Use the same equation, but use the following numbers instead:

a = acceleration due to gravity
v0 = velocity right before the ball is released
x0 = the position of the ball above the elevator floor (x = 0)

[Naeem]

Well, I did this :

x -x0 = vot + 1/2 at^2

a = -g = 9.81 m/s^2

v0 = 0 , the velocity of the ball before release is zero.

x0 = 0 , also equal to zero.

x - x0 = h = 2.1 m

So, pluggin in

t = 0.42 s , which is wrong!!!!

[jdavel]

Naeem,

It's t^2, not t.

[Naeem]

b) Now let's try a somewhat harder variation on this problem. Suppose instead that the elevator is accelerating upward with constant acceleration a = 1.2 m/s2 and that the upward speed of the elevator at the instant the ball is dropped, is 4.6 m/s. Now how long does it take the ball to hit the elevator floor, assuming once again that it is dropped from a height of 2.1 m?

Ok, I got part a) ... Now with this part,

I know, need to use the same equation, but how to plug in???

[asrodan]

Ball's position x = x0 + v0t - 1/2gt^2 = 2.1 + 4.6t - 1/2gt^2

Elevator's position y = y0 + v0t + 1/2at^2 = 4.6t + 1/2*1.2t^2

Solve for solve x = y for t

[mechanicsbook]

Imagine you have a camera inside and someone in a room is looking to a TV conected to it.He can not observe if the elevator is in repaos or has a constant speed .He can observe (is possible)only if it has acceleration. So ,in this situation,the time belongs to this equation :h=g*squaret:2 with h=1,8m g=10m/s2 .

[HallsofIvy]

b) Now let's try a somewhat harder variation on this problem. Suppose instead that the elevator is accelerating upward with constant acceleration a = 1.2 m/s2 and that the upward speed of the elevator at the instant the ball is dropped, is 4.6 m/s. Now how long does it take the ball to hit the elevator floor, assuming once again that it is dropped from a height of 2.1 m?

Ok, I got part a) ... Now with this part,

I know, need to use the same equation, but how to plug in???

You can add the two accelerations. The acceleration due to gravity, of the ball is 9.8 m/s2 (downward) and the elevator is accelerating upward at 1.2 m/s[sup]2[sup]. The ball is accelerating toward the floor of the elevator at 9.8+ 1.2= 11.0 m/s2. Again the initial speed of the elevator is irrelevant- the ball also has that initial speed.

1.2= (1/2)(11.0)t2.

[Naeem]

Yeah, I got it !!
Thanks,,