
#1
Oct3106, 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??? 



#2
Oct3106, 06:11 PM

P: 1,445

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?? 



#3
Oct3106, 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. 



#4
Oct3106, 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




#5
Oct3106, 06:18 PM

P: 147

But the apex is not the final velocity. It comes back down again.




#6
Oct3106, 06:18 PM

P: 1,445

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 



#7
Oct3106, 06:21 PM

P: 106

I get x= 1.637600545 is this right? (i am too stressed to round any numbers LOL)




#8
Oct3106, 06:22 PM

P: 1,445





#9
Oct3106, 06:23 PM

P: 147

That's not what I got.
What to you get for the time? 



#10
Oct3106, 06:25 PM

P: 106

i took 5.665418824(0.578103962)+0.5(9.8)(0.57810396)^2 and got 1.637600545




#11
Oct3106, 06:26 PM

HW Helper
P: 3,225

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. 



#12
Oct3106, 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 



#13
Oct3106, 06:33 PM

P: 1,445

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 



#14
Oct3106, 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




#15
Oct3106, 06:40 PM

HW Helper
P: 3,225

Btw, I suggest you do a few searches named 'projectile motion', this could do good. 



#16
Oct3106, 06:41 PM

Mentor
P: 4,499

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 



#17
Oct3106, 06:44 PM

P: 1,445

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 



#18
Oct3106, 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!!!!!!!!



Register to reply 
Related Discussions  
projectile motion range  Introductory Physics Homework  3  
Help! Maximizing Range of Projectile Motion!  Introductory Physics Homework  5  
Projectile Motion: Range  Introductory Physics Homework  3  
Projectile Motion: Range proof  Introductory Physics Homework  0 