f a ball dropped from a tower reaches the ground after 3.5 seconds, what is the height of the tower? Given: g = –9.8 meters/second2

2 Answer

  • Let's use the second kinematic equation:
    Δy = vot + 1/2gt²

    Δy = 0(3.5) + 1/2(-9.8)(3.5)²

    Δy = -4.9(3.5)² = -60.03 m

    So the ball falls 60.03 m, meaning that's the height of the building. The negative sign just means I put the y = 0 mark at the top of the building.

    Make sense?
  • The height of the tower from which a ball is dropped is 60.03 m.  

    The height of the tower can be calculated with the following equation:

    [tex] y_{f} = y_{i} + v_{i_{y}}t - \frac{1}{2}gt^{2} [/tex]


    [tex] y_{f} [/tex]: is the final height = 0 (when it reaches the ground)

    [tex] y_{i} [/tex]: is the initial height =?

    [tex] v_{i_{y}}[/tex]: is the initial velocity = 0 (the ball is dropped)

    t: is the time = 3.5 s

    g: is the acceleration due to gravity = 9.8 m/s²

    Hence, the height is:

    [tex] y_{i} = \frac{1}{2}9.8 m/s^{2}*(3.5 s)^{2} = 60.03 m [/tex]  

    Therefore, the height of the tower is 60.03 m.

    You can find another example of freefall here

    I hope it helps you!