#projectiles on slopes

16 messages · Page 1 of 1 (latest)

hybrid surge
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is it not 30?

light brook
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The maximum height reached by the ball thrown at 30 degrees from the hoizontal is solve using the formula ymax = vi^2 * (sin 30^2) / 2g where ymax is the maximum height in meters, vi^2 is the square of the initial velocity, and 2g is twice the value of the acceleration due to gravity or 19.6 m/s^2.

Solving for the maximum height, ymax

ymax = vi^2 * (sin 30^2) / 2g

ymax = (18m/s)^2 * (0.5)^2 / 19.6 m/s^2

ymax = 324 (m/s)^2 * 0.25 / 19.6 m/s^2

ymax = 4.1327 m

Now we can use sin(30) =4.1327/slope length
Slope length = 4.1327/0.5 = 8.2653

And then do 8.2653/18 = 0.46 so 0.46 seconds???

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Honestly I don’t even know

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I’m just waffling here

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I think it’s wrong tho

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Acc yh I am wrong I used the wrong formula

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this was the correct one

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where u is the initial velocity

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g is gravitational acceleration

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and theta is angle ofc

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So we get (2x18 x0.5)/9.8 which leaves us with 1.8s

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@harsh kestrel

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light brook
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Yh I just like to go ahead in terms of maths syllabus cos I'm picking maths further maths physics and computer science and/or chemistry which are all maths based

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so its better to be ahead