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    A ball is released from rest at height h metres above ground level. The ball hits the ground 1.5 seconds after it is released. Assume that the ball is a particle that does not experience any air resistance.

    A) Show that the speed of the ball is 14.7ms-1 when it hits the ground.

    I did v=0+(-9.8)(1.5)

    B) FInd H

    I did

    S=(0+14.7)/2 *1.5 then I timed the final answer by two to get 22.05m

    C) Find the distance that the ball has fallen when its speed is 5ms-1

    I got the answer as 14.775

    I did s = (u+v)/2 *t

    S = (5+14.7)/2 *1.5

    I considered the 5ms-1 as though it is positive despite the ball is desencding so that it does not have any bad effect on the distance. So I was wondering if it is right?

    S = ?
    U = 5ms-1
    V=14.7
    A=9.8
    T=1.5
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    (Original post by abdul95)
    A ball is released from rest at height h metres above ground level. The ball hits the ground 1.5 seconds after it is released. Assume that the ball is a particle that does not experience any air resistance.

    A) Find the distance that the ball has fallen when its speed is 5ms-1

    I got the answer as 14.775

    I did s = (u+v)/2 *t

    S = (5+14.7)/2 *1.5

    I considered the 5ms-1 as though it is positive despite the ball is desencding so that it does not have any bad effect on the distance. So I was wondering if it is right?

    S = ?
    U = 5ms-1
    V=14.7
    A=9.8
    T=1.5
    Sorry where did you get v=14.7 from?
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    (Original post by abdul95)
    A ball is released from rest at height h metres above ground level. The ball hits the ground 1.5 seconds after it is released. Assume that the ball is a particle that does not experience any air resistance.

    A) Find the distance that the ball has fallen when its speed is 5ms-1

    I got the answer as 14.775

    I did s = (u+v)/2 *t

    S = (5+14.7)/2 *1.5

    I considered the 5ms-1 as though it is positive despite the ball is desencding so that it does not have any bad effect on the distance. So I was wondering if it is right?

    S = ?
    U = 5ms-1
    V=14.7
    A=9.8
    T=1.5
    That would be the wrong equation to use for that as you would have two unknowns (s and t). You would be better off using v^2 = u^2 + 2as and rearranging it to make s the subject.
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    (Original post by abdul95)
    A ball is released from rest at height h metres above ground level. The ball hits the ground 1.5 seconds after it is released. Assume that the ball is a particle that does not experience any air resistance.

    A) Show that the speed of the ball is 14.7ms-1 when it hits the ground.

    I did v=0+(-9.8)(1.5)

    B) FInd H

    I did

    S=(0+14.7)/2 *1.5 then I timed the final answer by two to get 22.05m

    C) Find the distance that the ball has fallen when its speed is 5ms-1

    I got the answer as 14.775

    I did s = (u+v)/2 *t

    S = (5+14.7)/2 *1.5

    I considered the 5ms-1 as though it is positive despite the ball is desencding so that it does not have any bad effect on the distance. So I was wondering if it is right?

    S = ?
    U = 5ms-1
    V=14.7
    A=9.8
    T=1.5

    Part A is correct but accelleration would be positive 9.8 as the ball is falling to earth

    Part B, why have you multiplied final answer by 2?

    For part C, I would first find how long it takes the ball to reach the velocity of 5m/s then you can use the formula for displacement to calculate how far the ball has moved in this time. In part C the fact that the ball is moving at 14.7m/s when it hits the ground is irrelevant, so ignore that for this part of the question.
 
 
 
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