Question

An object is dropped from a bridge. A second object is thrown downward 1.48 s later. They both reach the water 48.1 m below at the same instant. What was the initial speed of the second object

Answer 1

To solve this problem we will apply the linear motion kinematic equations. With the data provided we will calculate the time of the first object to fall. Later we will get the time difference between the two. This difference will allow us to find the free fall distance. Through the distance we will find the initial velocity, that is,

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

$48.1 = 0*t + \frac{1}{2} (9.8)t^2$

$t = 3.13s$

The second object is thrown downward at one second later and it meets the first object at the water is

$t' = 3.13 -1.48$

$t' = 1.65s$

The distance of the object will travel due to free fall acceleration is

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

$x = 0*(1.65) +\frac{1}{2}(9.8)(1.65)^2$

$x = 13.34m$

The distance of the object will travel due to its initial velocity is

$v_0 = \frac{d_0}{t}$

$d_0 = v_0 t$

$48.1-13.34 = v_0 (1.65)$

$v_0 = 21.06m/s$

Therefore the initial speed of the second object is 21.06m/s