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soccerball3211
09-04-2006, 08:57 PM
At time t = 0 s, apple 1 is dropped from a bridge onto a roadway beneath the bridge; somewhat later, apple 2 is thrown down from the same height. Figure 2-25 gives the vertical positions y of the apples versus t during the falling, until both apples have hit the roadway. With approximately what speed is apple 2 thrown down?

apple 1 hits the ground 2 seconds after it is dropped
apple 2 is dropped at time t=1s and hits the ground 1.25 seconds later

I found the acceleration of apple 2 to be 25.13 m/s^2

I then calculated the initial velocity of apple 2 to be 31.4 m/s!

Could someone check my work?

stapel
09-04-2006, 09:08 PM
I found the acceleration of apple 2 to be 25.13 m/s^2
So these apples aren't being tossed around on this planet, since the acceleration (gravity) isn't -9.8 m/s<sup>2</sup>...?

Eliz.

soccerball3211
09-04-2006, 09:13 PM
final velocity = initial velocity + at

solve for initial velocity and you get 12.3 m/s

soccerball3211
09-04-2006, 09:18 PM
But 12.3 m/s is not the correct answer?

soccerball3211
09-04-2006, 09:32 PM
I know I made a mistake in my first attempt but I don't see where my mistake is when I calculated 12.3 m/s. Can someone help?

stapel
09-04-2006, 11:23 PM
We'll be glad to check your calculations, but you'll need to post your work for us to do that. Thank you.

Note: "Velocity" and "acceleration" are not the same thing. (I asked about the acceleration, and you replied about the velocity, is why I mention this.)

Eliz.

soccerball3211
09-04-2006, 11:44 PM
ok first I found the distance traveled by apple 1 by using the equation d=.5at^2.
d=.5at^2
d=.5(9.8m/s^2)(2s)^2
d=19.6m

then I used the equation final velocity = initial velocity + acceleartion multiplied by time
v sub f = 0
a =-9.8 m/s^2
t = 1.25 s

Therefore v sub i = 12.3 m/s

I also used the equation (v sub f)^2=(v sub i)^2 + 2 a d

and by using this equation I obtained a result of 19.6 m/s for the initial velocity

TchrWill
09-04-2006, 11:55 PM
At time t = 0 s, apple 1 is dropped from a bridge onto a roadway beneath the bridge; somewhat later, apple 2 is thrown down from the same height. Figure 2-25 gives the vertical positions y of the apples versus t during the falling, until both apples have hit the roadway. With approximately what speed is apple 2 thrown down?

apple 1 hits the ground 2 seconds after it is dropped
apple 2 is dropped at time t=1s and hits the ground 1.25 seconds later

I found the acceleration of apple 2 to be 25.13 m/s^2

I then calculated the initial velocity of apple 2 to be 31.4 m/s!

Could someone check my work.
apple 1 hits the ground 2 seconds after it is dropped
Thereforw, h = Vot + gt^2/2 = Vo(0) + 9.8(2^2)/2 = 19.6m.

apple 2 hits the ground 1.25 seconds after it is thrown.
Therefore, 19.6 = Vo(1.25) + 9.8(1.25^2)/2 = 7.65 making Vo = 9.555m/s

According to the times given. both apples do not hit the ground at the same time. Apple 2 is thrown from the 19.6m high bridge with a speed of 9.555m/s in order to hit the ground 1.25sec. later (.25 sec after the first apple hits the ground).

Did I go astray from the facts?

soccerball3211
09-05-2006, 12:17 AM
Or would the answer be 21.8 m/s
Because if you do 19.6m=Vi(1.25s)+.5(-9.8m/s^2)(1.25s)^2

Therefore Vi would be 21.8 m/s since the acceleration due to gravity is -9.8m/s^2

TchrWill
09-05-2006, 11:55 AM
Or would the answer be 21.8 m/s
Because if you do 19.6m=Vi(1.25s)+.5(-9.8m/s^2)(1.25s)^2

Therefore Vi would be 21.8 m/s since the acceleration due to gravity is -9.8m/s^2
Where did the .5 come from?

g = 9.8 is not negative when the object is falling from a height.

soroban
09-05-2006, 03:58 PM
Hello, soccerball3211!

At time t\,=\,0, apple-1 is dropped from a bridge onto a roadway beneath the bridge.
Somewhat later, apple-2 is thrown down from the same height.

Apple-1 hits the ground 2 seconds after it is dropped.
Apple-2 is thrown down at time t\,=\,1 . . . Where does it say that?
and hits the ground 1.25 seconds later

With what approximate speed is apple-2 thrown downward?

. . . And where does it say the apples hit the ground simultaneously?

The general equation for a free-fall problem is: \,y \:=\:h\,+\,vt\,-\,4.9t^2
. . where h = initial height, v = initial velocity.

With apple-1, v\,=\,0 (it is dropped).
Its height function is: \,y \:=\:h \,-\,4.9t^2

It hits the ground in 2 seconds: \,t\,=\,2,\;y\,=\,0
. . We have: 0 \:=\:h\,-\,4.9(2^2)\;\;\Rightarrow\;\;h\,=\,19.6 m.

Apple-2 has the height function: \,y \:=\:19.6\,+\,vt\,-\,4.9t^2
. . It hits the ground 1.25 seconds after it is thrown.
. . 0\:=\:19.6\,+\,v(1.25)\,-\,4.9(1.25^2)\;\;\Rightarrow\;\;v\:=\:-11.94375

Therefore, apple-2 is thrown downward at about 12 m/sec.

TchrWill
09-05-2006, 05:21 PM
At time t = 0 s, apple 1 is dropped from a bridge onto a roadway beneath the bridge; somewhat later, apple 2 is thrown down from the same height. Figure 2-25 gives the vertical positions y of the apples versus t during the falling, until both apples have hit the roadway. With approximately what speed is apple 2 thrown down?

apple 1 hits the ground 2 seconds after it is dropped
apple 2 is dropped at time t=1s and hits the ground 1.25 seconds later

I found the acceleration of apple 2 to be 25.13 m/s^2

I then calculated the initial velocity of apple 2 to be 31.4 m/s!

Could someone check my work?
Gravity slows down an object thrown upward with an initial velocity.

Gravity increases the velocity of an object thrown down with an initial velocity

Therefore:
apple 1 hits the ground 2 seconds after it is dropped
Thereforw, h = Vot + gt^2/2 = Vo(0) + 9.8(2^2)/2 = 19.6m, the height of the bridge.

apple 2 hits the ground 1.25 seconds after it is thrown downward from a height of 19.6m..
Therefore, 19.6 = Vo(1.25) + 9.8(1.25^2)/2 = 7.65 making Vo = 9.555m/s

The apple is thrown downward with an initial velocity of 9.555m/s.

soccerball3211
09-05-2006, 09:40 PM
Soroban it shows in the graph given that apple 2 is thrown down at time t=1 and lands on the ground at time t=2.25 thus landing 1.25 seconds after it is thrown downard.