Question

A key falls from a bridge that is 45 m above the water. the key falls straight down and lands in a model boat traveling at a constant velocity of 3.5 m/s. calculate the distance the boat was from the point of impact when the key was released.

Answer 1

Let the key is free falling, therefore from equation of motion

$h = ut +\frac{1}{2}gt^2.$.

Take initial velocity, u=0, so

$h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2$.

$h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2 \\\ t =\sqrt{\frac{2h}{g} }$

As velocity moves with constant velocity of 3.5 m/s, therefore we can use formula

$d= v \times t$

From above substituting t,

$d = v \times \sqrt{\frac{2h}{g} }$.

Now substituting all the given values and g = 9.8 m/s^2, we get

$d = 3.5 \ m/s \times \sqrt{\frac{2 \times 45 m}{9.8 m/s^2} } = 10.60 m$.

Thus, the distance the boat was from the point of impact when the key was released is 10.60 m.