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???
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??
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.
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
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.
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
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.
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
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
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!!!!!!!!
..since I'm an even nicer guy, I'll give you this Java applet to play with, and hopefully learn something from it: http://www.walter-fendt.de/ph11e/projectile.htm. {,:tongue2:}
correction SHE oooops k does the image work now?? look at the image. What is the final velocity in teh Y answer that question. THINK before you answer it And tell me WHY WHY WHYYYYYY you said that