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    'We know that we can assume gravity is constant and acting downwards, so we can use the
    constant acceleration relationships. Recall that a force resolved through 90o is zero – the same
    is true of velocities and acceleration. It acts at g m/s2 vertically downwards, but has zero
    acceleration component horizontally
    .'


    Can someone please explain this concept in a logical way..
    How can a= 0 horizontally and a =-g vertically ?

    I thought if u if different to v then there's some sort of accelaration be it -ve or +ve. Thanx in advance
    • Community Assistant
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    On the Earth, gravity acts downwards. It does not act sideways. The only force accelerating the projectile is gravity. By Newton's First Law, the projectile continues at constant speed in the horizontal direction as there is no force in that direction (zero acceleration in that direction as F = ma). It does not continue at constant speed in the vertical direction however as gravity is accelerating it towards the ground. Does that help?
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    It might help you to note that forces at right angles have no effect on each other. Think of acceleration as a vector.
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    (Original post by face05)
    Can someone please explain this concept in a logical way..
    How can a= 0 horizontally and a =-g vertically ?
    Because it is a vector.

    Draw a scale diagram of a right angled triangle (it will be slim) of width 0 and length 9.81 cm. The hypotenuse of the triangle is the resultant acceleration.
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    (Original post by Mr M)
    On the Earth, gravity acts downwards. It does not act sideways. The only force accelerating the projectile is gravity. By Newton's First Law, the projectile continues at constant speed in the horizontal direction as there is no force in that direction (zero acceleration in that direction as F = ma). It does not continue at constant speed in the vertical direction however as gravity is accelerating it towards the ground. Does that help?
    I thought the motion was caused by the initial force applied by the projectile to the projected particle :confused:

    I can see why a= 0 horizontally if a is caused by gravity ... but then how come a = -g .....

    I'm sorry my head is not getting anything today.

    Thank you all
    I dont understand it fully yet ... but I'm getting there

    So this new concept change the relationship between v, u , s and a

    Vx= Ux + AT... but a=0 therefor Vx = Ux = ucos( θ )

    Sx = ucos( θ )T

    How do you use the above equations to solve this question

    1. If a projectile is fired on a flat plain, at an angle of 40 deg to the horizontal with an initial velocity of 25m/s, how far away will the projectile land?

    all help appreciated ...thanx
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    You have to break down the question into two steps:
    a) After how long does the particle come back down to the ground?
    b) During that time, how long does it travel horizontally?

    For (a) you need to work with the vertical direction. Say that the initial velocity is u, which is at 40° to the horizontal, then the horizontal component of this velocity is u_x = u\cos 40^o and the vertical component is u_y = u\sin 40^o. Also, we have a downward vertical acceleration of g m/s², so the horizontal component of the acceleration is a_x = 0 and the vertical component is a_y = -g. Because of the nature of the acceleration, the time it takes for the particle to go all the way up to its maximum height is exactly half the time it takes for it to go up and come down (if we ignore air resistance and suchlike), so if we let T be the amount of time it takes to go up and down, then we can use the equation v_y = u_y + a_y\frac{T}{2}, where v_y = 0 (because its velocity is zero at its maximum height). You can then solve this for T to get your time.

    From there, part (b) is quite straightforward since a_x = 0, so s_x = u_xT, so you just sub in the values of u_x and T [which you found in part (a)] and you're done.

    Spoiler:
    Show
    EDIT: From this, you can derive a general formula which gives immediately what the horizontal distance a particle travels is if it is projected from a point with velocity u at angle \theta to the horizontal.

    Spoiler:
    Show
    \newline \displaystyle s_x =\dfrac{u^2\sin 2\theta}{g}
    • Thread Starter
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    Thought so... kinda knew you have to find T ....thanx anyway. Thank you all
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    It's negative because it's a vector quanitty so since it acts in the opposite direction which is upwards you use the negative sign (the downward acc=+9.81 , so you should immediatly expect that the upward acc. = -9.81)
    P.S: i think you should also know that at the highest point of the motion the object is instantaneously at rest as during it's motion it is accelerating upwards ( in other words , velocity is decreasing )
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    (Original post by nuodai)
    You have to break down the question into two steps:
    a) After how long does the particle come back down to the ground?
    b) During that time, how long does it travel horizontally?

    For (a) you need to work with the vertical direction. Say that the initial velocity is u, which is at 40° to the horizontal, then the horizontal component of this velocity is u_x = u\cos 40^o and the vertical component is u_y = u\sin 40^o. Also, we have a downward vertical acceleration of g m/s², so the horizontal component of the acceleration is a_x = 0 and the vertical component is a_y = -g. Because of the nature of the acceleration, the time it takes for the particle to go all the way up to its maximum height is exactly half the time it takes for it to go up and come down (if we ignore air resistance and suchlike), so if we let T be the amount of time it takes to go up and down, then we can use the equation v_y = u_y + a_y\frac{T}{2}, where v_y = 0 (because its velocity is zero at its maximum height). You can then solve this for T to get your time.

    From there, part (b) is quite straightforward since a_x = 0, so s_x = u_xT, so you just sub in the values of u_x and T [which you found in part (a)] and you're done.

    Spoiler:
    Show
    EDIT: From this, you can derive a general formula which gives immediately what the horizontal distance a particle travels is if it is projected from a point with velocity u at angle \theta to the horizontal.

    Spoiler:
    Show
    \newline \displaystyle s_x =\dfrac{u^2\sin 2\theta}{g}
    Meaning that is takes exactly the same amount of time to go up to its maxium height as it does to come back to its intial position , right?

    Thank you all

    Do you reckon that because Vy= 0 and sub that into

    Vy = Usin(theta) - gt = 0
    Usin(theta) = gt

    Usin(theta) /g = t

    I understand it now.
 
 
 
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