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 +(1)/(2)gt^2.`.

Take initial velocity, u=0, so

`h = 0* t + (1)/(2)g t^2= (1)/(2)gt^2`.

`h = 0* t + (1)/(2)g t^2= (1)/(2)gt^2 \\\ t =\sqrt{(2h)/(g) }`

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

`d= v * t`

From above substituting t,

`d = v * \sqrt{(2h)/(g) }`.

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

`d = 3.5 \ m/s * \sqrt{(2 * 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.