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    A stone is thrown vertically upwards at a speed of 16m/s from a point h metres above the ground. the stone hits the ground 4secs later.

    a) calculate the value of h

    b)the speed of the stone as it hits the ground


    HELLLP Please
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    am I stupid or do I need to know if it is thrown vertically upwards or downwards? anyway, you'll have to think about g when solving the problem.
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    oh yh vertically upwards sorry
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    For question 1

    s=ut + (at^2)/2

    (I think that formula will do...but there are other ones as well)

    For question 2

    V=u+at

    v=velocity
    u=initial velocity
    a=acceleration
    t=time
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    Take up as positive. The initial speed is 16m/s (u = 16), time is 4 seconds (t = 4), and the acceleration is 9.8m/s^2 down (a = -9.8). h is a distance. Do you know the formula connecting u, a, t, and distance (normally x or s)? For the second part - do you know a formula connecting u, a, t and the final speed (v), or can you read the above post and copy the formula? :p:
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    hmm still nt sure
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    (Original post by EconomyMan)
    hmm still nt sure
    Oh, I see. You want me to do your work for you.

    Not happening.
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    (Original post by generalebriety)
    Oh, I see. You want me to do your work for you.

    Not happening.
    its not homework...im just doing it for revision...if i get a modell answer...i know what im working towards.
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    (Original post by EconomyMan)
    its not homework...im just doing it for revision...if i get a modell answer...i know what im working towards.
    You try it. I will guide you along the way if that's what you want. And when you've done that, I will provide you with a model answer and more examples if you're still not understanding. But I'd like you to try this, since it is revision and not learning.
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    I am not fond of solving whole problems either...

    but I'll give you number 2...
    (In White text)
    v=u+at

    a=-10m/s^2
    t=4s
    u=16m/s

    hence v=-24m/s and as the question asks for speed and not velocity the answer is 24m/s
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    (Original post by nota bene)
    I am not fond of solving whole problems either...

    but I'll give you number 2...
    (In White text)
    v=u+at

    a=-10m/s^2
    t=4s
    u=16m/s

    hence v=-24m/s and as the question asks for speed and not velocity the answer is 24m/s
    Why must people do this? I really don't understand it. Posting entire solutions helps no one. It's a waste of everyone's time.
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    I don't like doing it as the OP will not learn anything, but I just did this one as I didn't have time to take another example... and still don't have... but I'll give one anyway and might help the OP then...

    A ball is thrown vertically upwards with a speed of 24.0m/s
    (a) when is the velocity of the ball 12m/s? and what is the displacement?
    (b) when is the velocity of the ball -12m/s? and what is the displacement?
    (c) what is the velocity of the ball 1.50 s after launch?
    (d) what is the maximum height reached by the ball?
 
 
 
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