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fucking suvat
Created 3rd October 2010 @ 13:36
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I am ok at them as long as they aren’t mega-hard, if you want tow rite question I can see if I can help.
either it’s hard or i’m having a senior moment.
question goes along these lines:
a man throws a basketball off a 30m high cliff at 34 degrees above the horizontal, if the horizontal range of the ball is 30m, what speed did he throw the ball?
i’m struggling on it, as far as i can see i only have one bloody variable. well, i have horizontal acceleration but that’s 0 so cancels out parts of the suvat equations.
Edit: might have an idea
Last edited by vlad_drac,
I didn’t know the term horizontal range applies to the distance reached by a projectile when it reaches its original height.
From the looks of it… it seems likely to be a simultaneous equation. I normally deal with them when you are given the inital horizontal velocity as well.
well, lets think about this logically. If we exclude air resistance, and it is in a purely parabolic motion.
Lets work out the vertical, so we know the time it takes to hit the ground
S-30m
U-0
V-?
A-9.8
T-?
now we use S=ut+1/2at^2
which gives us 30=1/2*9.8*t^2
rearrange equation then square root to get t
t=2.47
Now we can do the Horizontal
S-30
u-x
v-x
a-0
t-2.47
now we use S=ut+1/2at^2 again
S=xt+1/2t^2 (as a=0)
30=2.47x
x=12 to 2 sf
I’m really tired atm, so i’ve probably got it wrong, but that would be how you’d do it. Dead easy.
Last edited by Chaplain,
Quoted from octochris
0 == 1
lolthx Chris. I’ve written the Horizontal Accel as 0, not as 1, so i’ll re do it from there
now we use S=ut+1/2at^2 again
S=xt+1/2t^2 (as a=1)
30=2.47x + 1/2*2.47^2
x=10.91m/s
x(Initial Speed) to 2sf=11m/s
Lol thanks Chris, epic fail on my part there
Last edited by Chaplain,
Hard via text, i know, but:
– Lets work out the vertical, so we know the time it takes to hit the ground
– S-30m
– U-0
– V-?
– A-9.8
– T-?
Why have you started with initial vertical velocity as 0? The basketball starts with speed s at 53 degrees above the horizontal, so the initial vertical velocity is s * sin 53, and we don’t know what s is. If you take it from the top of its parabola, vertical speed is 0, but then we don’t know the height anymore.
Also in your edit, why is a=1? horizontal acceleration is 0, vertical acceleration is 9.18. So actually 1/2 at^2 would cancel out to 0.
—
Also, i’m gonna go back on myself on that one, i think the horizontal range in this case means when it hits the floor at the bottom of the cliff. There is a formula for uneven ground but you’re not gonna like it:
http://en.wikipedia.org/wiki/Range_of_a_projectile#Uneven_Ground
Believe me, sticking values into that gives you something like 3.5 million on one side of the equation, and v^4, v^2 and v on the other side.
Last edited by vlad_drac,
Quoted from vlad_drac
Hard via text, i know, but:
– Lets work out the vertical, so we know the time it takes to hit the ground
– S-30m
– U-0
– V-?
– A-9.8
– T-?Why have you started with initial vertical velocity as 0? The basketball starts with speed s at 53 degrees above the horizontal, so the initial vertical velocity is s * sin 53, and we don’t know what s is.
Also in your edit, why is a=1? horizontal acceleration is 0, vertical acceleration is 9.18
—
Also, i’m gonna go back on myself on that one, i think the horizontal range in this case means when it hits the floor at the bottom of the cliff. There is a formula for uneven ground but you’re not gonna like it:
http://en.wikipedia.org/wiki/Range_of_a_projectile#Uneven_Ground
Believe me, sticking values into that gives you something like 3.5 million on one side of the equation, and v^4, v^2 and v on the other side.
I originally had a=0, but I changed it due to Chris putting 0==1 or whatever, but I probably shouldn’t have. If you put the a=0, then the final answer is 12 m/s. The method is still correct, so just do it any way you like. i can’t be arsed to change the a back to the way it was in the first place.
I didnt actually read that he threw it at an angle, which was silly of me. You should draw a diagram to equalise it into Horizontal and Vertical speeds/forces and take it from there, continuing with the parabolic method I’ve done above (with a=0). If he threw it at a purely horizontally, then my method would be correct, so just equalise the forces with the xcos() method. I’m gonna go sleep, if you still havent got it later, then ill show you what i mean
I know what you mean, yeah. But do you know what i mean? :/
If he’s throwing it at an angle upwards, we don’t know the initial horizontal or vertical speed. We don’t know the time, we only know the horizontal distance (for uneven ground) and the vertical acceleration.
If you do a quick diagram yourself, i think you might see what i mean. Whether it’s vcos34 or vsign34, it’s still a proportion of a variable that we don’t know.
So basically, we know we know horizontal accel is 0, horizontal displacement is 30.
Last edited by vlad_drac,
horizontally:
30 = utcos(34)
vertically:
v = usin(34) + gt
Using the equation derived from the horizontal motion we see:
v = usin(34) + 30g/(ucos(34))
(v^2 = (usin34)^2 + 900g^2/(ucos34)^2 + 60gtan34)
using v^2 = u^2 + 2as we see:
v^2 = (usin34)^2 + 60g)
hence
60g = 60gtan(34) + 900g^2/(ucos34)^2
1-tan34 = 15g/(ucos34)^2
u=25.6ms^-1
this might be wrong i cba to check
Last edited by Zeks,
Quoted from Zeks
horizontally:
30 = utcos(34)vertically:
v = usin(34) + gt
Using the equation derived from the horizontal motion we see:
v = usin(34) + 30g/(ucos(34))
(v^2 = (usin34)^2 + 900g^2/(ucos34)^2 + 60gtan34)
using v^2 = u^2 + 2as we see:
v^2 = (usin34)^2 + 60g)hence
60g = 60gtan(34) + 900g^2/(ucos34)^2
1-tan34 = 15g/(ucos34)^2
u=25.6ms^-1this might be wrong i cba to check
You lost me a few steps along there, and the answer’s wrong too :(
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