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    A ball is launched from the top of a tower 35m high with an initial velocity 8.0ms-1 at an angle 25⁰ to the horizontal.
    Calculate:
    a) the maximum height reached by the ball above the top of the tower. [5]
    b) the time taken for the ball to reach the ground .[6]
    c) the range of the ball .[3]
    d) the magnitude and direction of the balls velocity at the moment of impact [7]

    So far I have resolved the initial velocity into vertical and horizontal components, now I'm not sure what the next step is. Any help is appreciated, thanks
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    Start by using the vertical component you have calculated to find the height.
    For the vertical motion you have
    initial velocity (vertical component)
    final velocity zero at maximum height
    acceleration is -g
    Use a suvat equation.
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    (Original post by rjnknr)
    A ball is launched from the top of a tower 35m high with an initial velocity 8.0ms-1 at an angle 25⁰ to the horizontal.
    Calculate:
    a) the maximum height reached by the ball above the top of the tower. [5]
    b) the time taken for the ball to reach the ground .[6]
    c) the range of the ball .[3]
    d) the magnitude and direction of the balls velocity at the moment of impact [7]

    So far I have resolved the initial velocity into vertical and horizontal components, now I'm not sure what the next step is. Any help is appreciated, thanks
    usually for projectile there are special formulae used to calculate things such as range, max height, ...etc.

    but I still have problems using these formulae. However I realized that these formulae are derivatives of the suvat equations. so what I did was using the suvat equations to solve such questions.

    by using suvat:

    (a) consider the vertical motion of the particle to find the max height from the point of projection. you can do that by suvat since you already found the vertical component of the initial velocity, you know the final velocity which must be equal to 0 and the acceleration which is g=9.81. the answer you'll find by the previous calculations will be enough, but when asked to find the max height from the ground you should add the length of the tower to the value of the max height from point of projection.

    (b) you can do that again by considering the vertical motion, since you have the vertical component of initial velocity, distance to ground (length of tower), and acceleration (g=9.81).

    (c) you must consider the horizontal motion here, and since the horizontal velocity is constant, there will be no acceleration (a=0), so you'll be using s=ut . you do have the horizontal component of the initial velocity and you would have found the time in part (b) .

    (d) in that part you must find both the vertical and horizontal components of final velocity for the particle, but since the horizontal velocity is constant, the final velocity will be equal to the initial velocity when it comes to the horizontal component. that make things easier in this part, but you still have to work out the vertical component which you can find in a similar way to that used in part (b). however here you'll be working to find the final velocity instead of time. then you'll have to get the magnitude of the velocity by applying Pythagoras' theorem: root[vertical2+horizontal2]. the direction can be worked out by basic trigonometry: angle=tan(vertical/horizontal).

    *note: you might want to set a positive sense before solving such questions to differentiate between positive and negative values because this is a common mistake. vectors with directions against your positive sense must be negative.

    that's how you can tackle such questions using suvat equations, but if you're using an alternative method I hope someone else could help you because this is the only method I know

    I hope that was helpful
 
 
 
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