2 apples: With approximately what speed is it thrown?

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soccerball3211

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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?
 
soccerball3211 said:
I found the acceleration of apple 2 to be 25.13 m/s^2
Click to expand...
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.
 
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?
 
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.
 
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
 
soccerball3211 said:
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.
Click to expand...
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?
 
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
 
soccerball3211 said:
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
Click to expand...
Where did the .5 come from?

g = 9.8 is not negative when the object is falling from a height.
 
Hello, soccerball3211!

At time \(\displaystyle t\,=\,0\), 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 \(\displaystyle t\,=\,1\) . . . 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?
Click to expand...

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

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

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


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

Therefore, apple-2 is thrown downward at about \(\displaystyle 12\) m/sec.

 
soccerball3211 said:
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?
Click to expand...
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.
 
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.
 
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