Physics

A stone is thrown vertically upwards from the top of a tower 50.0m high with an intial velocity of 20.0m/s. (a) whats the maximum height the stone reaches? (b) Whats the time it takes to reach the maximim height? (c) What is the total distance it covers?

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  1. vf^2=2gh where h is the height from the tower. for total height, add 50m

    time? h=hi+vi*t-4.9t^2
    solve for t

    distance? 2h+50

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    bobpursley
  2. Initial velocity=20 m/s
    Final velocity=0m/s
    Height using third equation of motion 2as=v²-u²
    Acceleration due to gravity = -9.8 m/s(-10m/s)
    - sign for acceleration in opposite direction
    2*-10*s=0²-20²
    -20s=-400
    S=-400/-20=20metres
    So maximum height= height of tower + height of stone from tower
    50+20=70 m
    b) time taken using first equation of motion v=u+at
    0=20+(-10t)
    -20=-10t
    t=-20/-10 =2 seconds

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