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borishaifa [10]
3 years ago
11

An object accelerates from rest to 85m/s over a distance of 36m. What acceleration did it experience?

Physics
1 answer:
Marizza181 [45]3 years ago
6 0
Suvat
we have s, u, v and we want a
the suvat equation with these values in is: v^2 = u^2 - 2as
so a = (-v^2 + u^2)/-2s 
plug numbers in
a = (-85^2 + 0^2)/-2*36 = 7225/72 = 100.3... ms^-2
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Allushta [10]

Answer:

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          x = v₀ₓ t

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          45 = v₀ cos θ t

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           v_{y} = v_{oy} - gt

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         45 = v₀ cos θ  t

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         0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

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we substitute

        10 = (9.8 t) t - ½ 9.8 t2

        10 = ½ 9.8 t2

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Now let's use the first equation and the last one

         45 = v₀ cos θ t

         0  = v₀ sin θ - 9.8 t

         9.8 t = v₀  sin θ

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we divide

         9.8t / (45 / t) = tan θ

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Now we can calculate the maximum height

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the other angle that gives the same result is

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for this angle the maximum height is

 

          y = v_{oy}^2 / 2g

          y = (20 sin 66)²/2 9.8

           y = 17 m

thisis the correct

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