# projectile motions help

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#1
I tried using Pythagoras and had ended up with 20.396 m/s but the given answer is different. Please can someone help me? (See image)
Last edited by ddsizebra; 3 years ago
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3 years ago
#2
(Original post by ddsizebra)
I tried using Pythagoras and had ended up with 20.396 m/s but the given answer is different. Please can someone help me? (See image)
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#3
(Original post by mqb2766)
Can you show you're working?

(Original post by mqb2766)
Can you show you're working?

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3 years ago
#4
(Original post by ddsizebra)
I'm not sure why you are using pythagoras for the vertical and horizontal distances...? Those are not the horizontal and vertical components of the velocity so you can't perform pythagoras on those values to calculate the velocity.

Instead, you should consider the motion of the ball in the vertical and horizontal directions and use the relevant SUVAT equations.
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3 years ago
#5
The velocities are not right.
Use SUVAT to work out the initial vertical velocity and the time taken to reach the top of the wall, then use that to find the horizontal velocity. Then combine.
(Original post by ddsizebra)
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#6
(Original post by mqb2766)
The velocities are not right.
Use SUVAT to work out the initial vertical velocity and the time taken to reach the top of the wall, then use that to find the horizontal velocity. Then combine.
but velocity or time was given let alone the angle. I just don't know where to start...
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3 years ago
#7
You don't need the angle, horizontal and vertical are independent. Combine using Pythagoras.

Use SUVAT to find initial velocities and time.
Last edited by mqb2766; 3 years ago
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3 years ago
#8
(Original post by ddsizebra)
but velocity or time was given let alone the angle. I just don't know where to start...
You can use energy:

Spoiler:
Show
0.5mU^2 = mgh + 0.5m(Ux)^2 (1) where h = 4 (Ux is constant throughout the motion assuming no air resistance)

Ux = Ucosθ -----> substituting this into (1) results in Usinθ = Uy = √(2gh)

Using suvat in the vertical direction gives Uy = gT so T = Uy/g = √(2h/g)

Ux = 20/T = 20 / √(2h/g)

so U = (Ux)^2 + (Uy)^2

You can do this just using suvat too if u want
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3 years ago
#9
(Original post by ddsizebra)
but velocity or time was given let alone the angle. I just don't know where to start...
If you are not happy with the concept of energy, you can do it entirely using suvat - it is much simpler.

(please have a genuine go before looking below):

Spoiler:
Show
In the horizontal direction, Ux = 20/T where T = time taken for ball to reach wall

In the vertical direction using v=u+at, Uy = gT and using v^2 = u^2 + 2as , (Uy)^2 = 2gh meaning Uy = sqrt(2gh)

Hence T = Uy / g = sqrt(2h/g) ----> Ux = 20/ sqrt(2h/g)

And of course finally U^2 = (Ux)^2 + (Uy)^2
1
#10
(Original post by Anonymouspsych)
If you are not happy with the concept of energy, you can do it entirely using suvat - it is much simpler.

(please have a genuine go before looking below):

Spoiler:
Show
In the horizontal direction, Ux = 20/T where T = time taken for ball to reach wall

In the vertical direction using v=u+at, Uy = gT and using v^2 = u^2 + 2as , (Uy)^2 = 2gh meaning Uy = sqrt(2gh)

Hence T = Uy / g = sqrt(2h/g) ----> Ux = 20/ sqrt(2h/g)

And of course finally U^2 = (Ux)^2 + (Uy)^2
3rd line looks very confusing...I tried rearranging as if simultaneous equation but struggled....
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3 years ago
#11
(Original post by ddsizebra)
3rd line looks very confusing...I tried rearranging as if simultaneous equation but struggled....
Ok so I am assuming you understood the 2nd line where I derived the two following equations:

Uy = sqrt(2gh) (1)
Uy = gT (2)

Substituting equation (2) in equation (1) gives gT = sqrt(2gh)
And dividing both sides by g gives T = sqrt(2h/g) (3) lets call this equation 3

Now on the first line I mentioned Ux = 20/T but now we know what T is from equation (3) so we can substitute it giving:
Ux = 20/T = 20/sqrt(2h/g)
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#12
(Original post by Anonymouspsych)
Ok so I am assuming you understood the 2nd line where I derived the two following equations:

Uy = sqrt(2gh) (1)
Uy = gT (2)

Substituting equation (2) in equation (1) gives gT = sqrt(2gh)
And dividing both sides by g gives T = sqrt(2h/g) (3) lets call this equation 3

Now on the first line I mentioned Ux = 20/T but now we know what T is from equation (3) so we can substitute it giving:
Ux = 20/T = 20/sqrt(2h/g)
but there is no gravity acceleration in X direction. Even if there is the result would be 22.136m/s (Not correct answer)

Is there a way to calculate the angle using the 2 distances via Pythagoras and Trigonometry? because what I have as theta is 11.31 degrees.
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3 years ago
#13
You should get
Uy = 8.85
T = 0.9
Ux = 22.2
(Original post by ddsizebra)
but there is no gravity acceleration in X direction. Even if there is the result would be 22.136m/s (Not correct answer)

Is there a way to calculate the angle using the 2 distances via Pythagoras and Trigonometry? because what I have as theta is 11.31 degrees.
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#14
(Original post by mqb2766)
You should get
Uy = 8.85
T = 0.9
Ux = 22.2
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3 years ago
#15
Quite a bit is wrong especially using the angle/trigonometry., Please use the suggestions earlier in the thread to
1) work out the initial vertical velocity
2) work out the time to reach the peak
3) work out the horizontal velocity
4) use Pythagoras to combine 1) and 3).

1) 2) and 3) all involve relatively simple suvat equations. Can you determine which ones the are and reply?

(Original post by ddsizebra)
Last edited by mqb2766; 3 years ago
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#16
(Original post by mqb2766)
Quite a bit is wrong, please use the suggestions earlier in the thread to
1) work out the initial vertical velocity
2) work out the time to reach the peak
3) work out the horizontal velocity
4) use Pythagoras to combine 2) and 4)

1) 2) and 3) all involve relatively simple suvat equations. Can you determine which ones the are and reply?
I've been pointing this out yesterday, saying that I cannot find initial velocity if final velocity, time, angle weren't given. And since initial velocity is made up of 2 components (vertical and horizontal), one of 2 components uses gravity as acceleration.

I'm really confused and have tried to use ALL SUVAT equations but that obviously failed. I'm just waiting on my lecturer see if he can explain it clearer....
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3 years ago
#17
Consider the initial vertical velocity. The
* acceleration is -9.8
* Distance travelled is 4
* Final velocity is 0
* Initial velocity is Uy
Which SUVAT equation relates these 4 terms and hence means you can calculate Uy?
(Original post by ddsizebra)
I've been pointing this out yesterday, saying that I cannot find initial velocity if final velocity, time, angle weren't given. And since initial velocity is made up of 2 components (vertical and horizontal), one of 2 components uses gravity as acceleration.

I'm really confused and have tried to use ALL SUVAT equations but that obviously failed. I'm just waiting on my lecturer see if he can explain it clearer....
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#18
(Original post by mqb2766)
Consider the initial vertical velocity. The
* acceleration is -9.8
* Distance travelled is 4
* Final velocity is 0
* Initial velocity is Uy
Which SUVAT equation relates these 4 terms and hence means you can calculate Uy?
using v^2 = u^2 + 2as

for which I got 8.854m/s

but can I assume final velocity =0 in horizontal direction too? and there would be not gravity acceleration so a=0m/s^2?
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3 years ago
#19
That looks right.
2) How long did it take to get there? Use a similar, but different SUVAT equation, again in the vertical direction only.
Forget about the horizontal direction until you've found the time.
(Original post by ddsizebra)
using v^2 = u^2 + 2as

for which I got 8.854m/s

but can I assume final velocity =0 in horizontal direction too? and there would be not gravity acceleration so a=0m/s^2?
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#20
(Original post by mqb2766)
That looks right.
2) How long did it take to get there? Use a similar, but different SUVAT equation?
Forget about the horizontal direction until you've found the time.
using uy=8.854m/s
and a=-9.81m/s^2
and vy=0ms

time would be 0.9025s. this is extremely small...
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