TSR George
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A ball is thrown at an angle 45 degrees to the horizontal. With what initial speed should the ball be thrown to land 100m away?
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S.G.
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Isn’t this just a projectile motion question?

Resolve vertically and perpendicular first. I haven’t done maths in ages but I guess from there you’d find values somehow.

Horizontally the initial speed is xcos45.
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Deranging
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What equations do you know?

Suppose that the initial speed is  u , what is the initial speed in the horizontal and the vertical direction?
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uberteknik
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(Original post by S.G.)
Isn’t this just a projectile motion question?

Resolve vertically and perpendicular first. I haven’t done maths in ages but I guess from there you’d find values somehow.

Horizontally the initial speed is xcos45.
(Original post by Deranging)
What equations do you know?

Suppose that the initial speed is  u , what is the initial speed in the horizontal and the vertical direction?
The question is asking for the initial unresolved vector speed to achieve a horizontal distance of 100m.

i.e. the time taken for the ball to reach the apex and then fall to the ground, must be equal to the time taken to cover the 100m distance.
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thotproduct
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got a bit curious and found this thread, might be a bit late but i decided to give this q a go myself, do tell if i made a mistake loool

so you're looking for some x, which is the magnitude of your velocity, it's at an angle of 45 deg to the horizontal, so it's obvs gonna have horizontal and vertical components, let's call those xcos45 and xsin45 respectively.

We know one thing immediately, it's total time of flight covers 100m, and the horizontal component is assumed to be constant.

So xcos45 * t = 100, so t = 100/xcos45.

We know that this time is going to be equal to the time it takes for the thing, vertically, to go from initial position to maximum height, and back down to ground again. We can suvat this batty boy vertically

s = we don't give
u = xsin45
v = 0 (at max height)
a = -9.8
t = ?

v = u + at

0 = xsin45 -9.8t.

so obvs 9.8t = xsin45 and t = xsin45/9.8, note that this is only the first half of the journey, the total will be double this, so 2xsin45/9.8 is total time

Set this equal to the time we got initially.

2xsin45/9.8 = 100/xcos45

Note that sin and cos 45 are both root2/2

gonna be a bit btec here and clean up LHS and RHS individually don't @ me

2x(root2/2)/9.8 which is xroot2/9.8

multiply through by 9.8 so

xroot2 = 980/xcos45

xroot2 * xcos45 = 980

root 2 * cos45 = 1

so x^2 = 980

x = 14root5

Is this right? This is how I'd go about it :rofl:
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(Original post by thotproduct)
got a bit curious and found this thread, might be a bit late but i decided to give this q a go myself, do tell if i made a mistake loool

so you're looking for some x, which is the magnitude of your velocity, it's at an angle of 45 deg to the horizontal, so it's obvs gonna have horizontal and vertical components, let's call those xcos45 and xsin45 respectively.

We know one thing immediately, it's total time of flight covers 100m, and the horizontal component is assumed to be constant.

So xcos45 * t = 100, so t = 100/xcos45.

We know that this time is going to be equal to the time it takes for the thing, vertically, to go from initial position to maximum height, and back down to ground again. We can suvat this batty boy vertically

s = we don't give
u = xsin45
v = 0 (at max height)
a = -9.8
t = ?

v = u + at

0 = xsin45 -9.8t.

so obvs 9.8t = xsin45 and t = xsin45/9.8, note that this is only the first half of the journey, the total will be double this, so 2xsin45/9.8 is total time

Set this equal to the time we got initially.

2xsin45/9.8 = 100/xcos45

Note that sin and cos 45 are both root2/2

gonna be a bit btec here and clean up LHS and RHS individually don't @ me

2x(root2/2)/9.8 which is xroot2/9.8

multiply through by 9.8 so

xroot2 = 980/xcos45

xroot2 * xcos45 = 980

root 2 * cos45 = 1

so x^2 = 980

x = 14root5

Is this right? This is how I'd go about it :rofl:
If you're a fan of memorising formulae, the easiest way to do this question would be to put it into

 y = x \text{tan}(\theta) - \dfrac{gx^2}{2u^2 \text{cos}^{2} (\theta)}

Which is essentially derived from resolving vertically and horizontally and combining them.
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