Question

A man drops a baseball from the top of a building. If the ball is held at a height of 1m before it is dropped, and takes 6.8 seconds to hit the ground, how high is the building in meters? (Neglect air resistance)

Answer 1

To solve this problem we will apply the linear motion kinematic equations.

The equation that describes the position as a function of the initial velocity, acceleration and time is given by the relation

$s = v_0 t +\frac{1}{2} at^2$

Here,

$v_0 =$ Initial velocity

t = Time

a = Acceleration, at this case due to gravity

There is not initial velocity then we have that the equation to the given time is

$s = \frac{1}{2} (9.8)(6.8)^2$

$s = 226.8m$

If the ball is held at a height of 1m before it is dropped, we have that the Building height is

$h = 226.8-1$

$h = 225.8m$