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Nov 3 2009 09:39pm
1.) You took a running leap off a high-diving platform. You were running at 3.1 m/s and hit the water 2.8 s later. How high was the platform, and how far from the edge of the platform did you hit the water? Ignore air resistance.
height of platform ___________ m
distance from the edge of the platform ___________ m


2. ) An arrow is shot at 27.0° above the horizontal. Its velocity is 45 m/s, and it hits the target.
(a) What is the maximum height the arrow will attain?
___________m

(b) The target is at the height from which the arrow was shot. How far away is it?
___________ m

3.) A pitched ball is hit by a batter at a 45° angle and just clears the outfield fence, 102 m away. If the fence is at the same height as the pitch, find the velocity of the ball when it left the bat. Ignore air resistance.
__________ m/s (at 45°)

4.) A player kicks a football from ground level with an initial velocity of 27.0 m/s, 30.0° above the horizontal, as shown below. Find the ball's hang time, range, and maximum height. Assume air resistance is negligible. (Figure is not to scale.)

Click Here For Figure http://www.webassign.net/gphys/7-08.gif

hang time ______s
range __________ m
maximum height ________ m

5.) A player kicks a football from ground level with an initial velocity of 27.0 m/s, 60.0° above the horizontal, as shown below. Find the ball's hang time, range, and maximum height. Assume air resistance is negligible. (Figure is not to scale.)
(Uses Same figure as number 4)

hang time ______s
range ______ m
maximum height ______ m



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Nov 3 2009 09:48pm
Quote (GladiatorZ_BG @ Nov 3 2009 10:39pm)
1.)  You took a running leap off a high-diving platform. You were running at 3.1 m/s and hit the water 2.8 s later. How high was the platform, and how far from the edge of the platform did you hit the water? Ignore air resistance.
height of platform ___________ m
distance from the edge of the platform ___________ m

h = 4.9(2.8^2)
h = 38.416
distance = 3.1(2.8)
distance = 8.68
2. )  An arrow is shot at 27.0° above the horizontal. Its velocity is 45 m/s, and it hits the target.
(a) What is the maximum height the arrow will attain?
___________m

(b) The target is at the height from which the arrow was shot. How far away is it?
___________ m

break up into vertical/horizontal components.
h = Vy*t - 4.9t^2. solve for t to find how long in air.
divide that t by two and plug back into height formula to find max height.
to find how far away target is:
distance = Vx*t

3.)  A pitched ball is hit by a batter at a 45° angle and just clears the outfield fence, 102 m away. If the fence is at the same height as the pitch, find the velocity of the ball when it left the bat. Ignore air resistance.
__________ m/s (at 45°)

basically its a relationship between Vx and Vy. vx = Vy since its at a 45 degree angle. and 102 = Vx*t. also 0 = Vxt- 4.9t^2. solve for t. after you find Vx square it to find the velocity at 45 degree angle.

4.)  A player kicks a football from ground level with an initial velocity of 27.0 m/s, 30.0° above the horizontal, as shown below. Find the ball's hang time, range, and maximum height. Assume air resistance is negligible. (Figure is not to scale.)

Click Here For Figure  http://www.webassign.net/gphys/7-08.gif

hang time  ______s
range  __________ m
maximum height  ________ m

do what i showed you in the second one.

5.)  A player kicks a football from ground level with an initial velocity of 27.0 m/s, 60.0° above the horizontal, as shown below. Find the ball's hang time, range, and maximum height. Assume air resistance is negligible. (Figure is not to scale.)
(Uses Same figure as number 4)

hang time ______s
range ______ m
maximum height ______  m

do what i showed in second one.


that should help.
/e ya ima have to some what agree with nickey. i can understand you not getting these problems once or twice, but you've posted homework similar to this for multiple days. and you have, what seems like, no intent to doing your homework yourself/learning.

This post was edited by cialda on Nov 3 2009 10:16pm
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Nov 3 2009 10:11pm
U NEED HELP ON THESE PROBLEMS?

...
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Nov 3 2009 10:16pm
Quote (nickey @ 3 Nov 2009 22:11)
U NEED HELP ON THESE PROBLEMS?

...


i already gave him the answers for the first questions expalined how to do them and left the rest for him, without a fg offering i dont see why anyone should do his hw for him since i already expalined, i said ask if you dont understand something, not post your hw again to see if someone esle will do the whole thing for you
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Nov 3 2009 10:31pm
Quote (cialda @ Nov 4 2009 03:48am)
that should help.
/e ya ima have to some what agree with nickey. i can understand you not getting these problems once or twice, but you've posted homework similar to this for multiple days. and you have, what seems like, no intent to doing your homework yourself/learning.


thats funny....

no intent?

fact is the similar homework i posted noone explained how to do.... all they did was post answeres.......
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Nov 3 2009 10:33pm
Quote (ass666 @ Nov 4 2009 04:16am)
i already gave him the answers for the first questions expalined how to do them and left the rest for him, without a fg offering i dont see why anyone should do his hw for him since i already expalined, i said ask if you dont understand something, not post your hw again to see if someone esle will do the whole thing for you


u did this?


dont remembe u solving any answeres or explaing them how to do.......... so dont post bs if ur lieing
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Nov 3 2009 10:40pm
Quote (cialda @ Nov 4 2009 03:48am)
that should help.
/e ya ima have to some what agree with nickey. i can understand you not getting these problems once or twice, but you've posted homework similar to this for multiple days. and you have, what seems like, no intent to doing your homework yourself/learning.




where did u even get 4.9 from?


1.) You took a running leap off a high-diving platform. You were running at 3.1 m/s and hit the water 2.8 s later. How high was the platform, and how far from the edge of the platform did you hit the water? Ignore air resistance.
height of platform ___________ m
distance from the edge of the platform ___________ m

h = 4.9(2.8^2)
h = 38.416
distance = 3.1(2.8)
distance = 8.68
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Nov 3 2009 10:49pm
Quote (cialda @ Nov 4 2009 03:48am)
that should help.
/e ya ima have to some what agree with nickey. i can understand you not getting these problems once or twice, but you've posted homework similar to this for multiple days. and you have, what seems like, no intent to doing your homework yourself/learning.


(b) The target is at the height from which the arrow was shot. How far away is it?
___________ m

break up into vertical/horizontal components.
h = Vy*t - 4.9t^2. solve for t to find how long in air.
divide that t by two and plug back into height formula to find max height.
to find how far away target is:
distance = Vx*t

Here not sure where your getting 4.9t^2.
Also would be nice if u can show the way u solve for t in this one
Also for distance Vx*t whats the Vx?

Talking about this problem


2. ) An arrow is shot at 27.0° above the horizontal. Its velocity is 45 m/s, and it hits the target.
(a) What is the maximum height the arrow will attain?
___________m

(b) The target is at the height from which the arrow was shot. How far away is it?
___________ m

break up into vertical/horizontal components.
h = Vy*t - 4.9t^2. solve for t to find how long in air.
divide that t by two and plug back into height formula to find max height.
to find how far away target is:
distance = Vx*t

This post was edited by GladiatorZ_BG on Nov 3 2009 10:49pm
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Nov 3 2009 10:52pm
basically its a relationship between Vx and Vy. vx = Vy since its at a 45 degree angle. and 102 = Vx*t. also 0 = Vxt- 4.9t^2. solve for t. after you find Vx square it to find the velocity at 45 degree angle.


Need to know how to solve for t for this one, also need to know how to Vx square it to find the velocity at 45 degree angles... dont have these formluas, i think ur using diff formulas then the 1s i was given

the formulas i was given are

y= 1/2gt2
t= squared root 2y/g
x= vxt
Vy= gt

This post was edited by GladiatorZ_BG on Nov 3 2009 10:55pm
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Nov 3 2009 10:56pm
4.9=0.5g=0.5*9.8

cmon
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