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Projectile Motion Range (please Help)

by Cole07
Tags: motion, projectile, range
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Cole07
#1
Oct31-06, 06:05 PM
P: 106
Question: How far will a stone travel over level ground if it is thrown upward at an angle of 31.0 degrees with respect to the horizontal and with a speed of 11.0m/s?


What is the maximum range that could be achieved with the same initial speed?


I have tried to solve this with no sucess what i have done is Drawn a triangle with a angle being 31.0 degrees the opposite side from the angle is Vy the adjacent side is Vx the hypothesis is 11.0m/s then i drew a chart with X and Y X side of the chart I have Vx= 9.428840308 a=0 then on the y side i have
Vy=5.665418824 a=-9.8 the Vf=0m/s I used the formula Vf=Vi+At and solved for time and got
0.578103962 then i used d=Vi*t+1/2At^2 and solved for D and got 1.637620735 for the first question (But was told this was not correct)

For the second question i doubled the time and multiplied by Vx D=9.428840308 (1.15620784) and got 10.90169909 but it was wrong !

Is there anyone who might be able to help me Please???
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stunner5000pt
#2
Oct31-06, 06:11 PM
P: 1,440
what is the final velocity in the Y direction? And why?

for your second question how do you maximize the distance you could travel? They gave you a hint : the SPEED is thesame but something else must be different. What must change?
and Why did you double time??
rsk
#3
Oct31-06, 06:12 PM
P: 147
GOod grief you don't believe in rounding numbers do you????

If you've calculated, as it appears, the time taken for the vertical velocity to reach 0M/s, then you've calculated the time for the stone to reach it's maximum height, - it still has some way to go before it hits the ground.

I would use here the equation x = V(i)t + 0.5 a t^2
with the vertical velocity and a = 9.8 but remembering that when the stone hits the ground its vertical displacement is zero because it started from ground level

That will give you a time, then use this with the horizontal velocity (constant) to calculate the horizontal distance.

Cole07
#4
Oct31-06, 06:14 PM
P: 106
Projectile Motion Range (please Help)

final velocity in y direction is 0 because at the apex it is not going anywhere
rsk
#5
Oct31-06, 06:18 PM
P: 147
But the apex is not the final velocity. It comes back down again.
stunner5000pt
#6
Oct31-06, 06:18 PM
P: 1,440
Quote Quote by Cole07
final velocity in y direction is 0 because at the apex it is not going anywhere
tahts fine

but is that the final velocity in the y direction for the WHOLE trip

after all you are trying to find the time it took for the ENTIRE journey
Cole07
#7
Oct31-06, 06:21 PM
P: 106
I get x= 1.637600545 is this right? (i am too stressed to round any numbers LOL)
stunner5000pt
#8
Oct31-06, 06:22 PM
P: 1,440
Quote Quote by Cole07
I get x= 1.637600545 is this right? (i am too stressed to round any numbers LOL)
howd you get it?
rsk
#9
Oct31-06, 06:23 PM
P: 147
That's not what I got.

What to you get for the time?
Cole07
#10
Oct31-06, 06:25 PM
P: 106
i took 5.665418824(0.578103962)+0.5(-9.8)(0.57810396)^2 and got 1.637600545
radou
#11
Oct31-06, 06:26 PM
HW Helper
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P: 3,224
Quote Quote by rsk
That's not what I got.

What to you get for the time?
Did you use [tex]y(t) = 11\cdot\sin(31)t-\frac{1}{2}gt^2=0[/tex] to calculate the time?

If you didn't, do so. (Solve for t.) When you do so, plug that time into [tex]x(t)=11\cdot\cos(31)t[/tex] to get the total displacement.
Cole07
#12
Oct31-06, 06:30 PM
P: 106
to get the time i used the formula Vf=Vi+at
Vf=0m/s
a= -9.8
Vi=5.665418824 (by 11.0sin(31.0))
and then i solved for time
stunner5000pt
#13
Oct31-06, 06:33 PM
P: 1,440
Quote Quote by Cole07
to get the time i used the formula Vf=Vi+at
Vf=0m/s
a= -9.8
Vi=5.665418824 (by 11.0sin(31.0))
and then i solved for time
why are you still saying velocity final in the Y is zero? Its NOT zero

after it has completed its entire flight what is the velocity in the Y just before it hits the ground? I certainly hope it isnt zero otherwise whatever ive been doing for these 6 odds years has been in vain! VAIN!

and this world has no justice
Cole07
#14
Oct31-06, 06:37 PM
P: 106
I don't understand? I am going by the example we did in class and for Vf we used 0 and this is at the apex
radou
#15
Oct31-06, 06:40 PM
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P: 3,224
Quote Quote by Cole07
I don't understand? I am going by the example we did in class and for Vf we used 0 and this is at the apex
Interesting class, indeed.

Btw, I suggest you do a few searches named 'projectile motion', this could do good.
Office_Shredder
#16
Oct31-06, 06:41 PM
Emeritus
Sci Advisor
PF Gold
P: 4,500
That would be good to calculate the maximum HEIGHT of the object, not the maximum distance it travels.

Throw a ball into the air. When it hits the ground, is the y velocity zero? Use intuition, it really helps sometimes
stunner5000pt
#17
Oct31-06, 06:44 PM
P: 1,440
since im such a nice guy i drew a diagram for you
look at the diagram

now tell me what do u think the velocity i nteh Y direction must be after thw whole trip?

THINK before you answer

but dont think too hard

you are probably mixing up the type of questions
Attached Thumbnails
1.JPG  
Cole07
#18
Oct31-06, 06:49 PM
P: 106
yes you are a nice guy but not only am i having trouble with physics BUT I CAN'T OPEN THE ATTACHMENT THIS IS SO SAD!!!!!!!!


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