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Question: An object falls from a bridge that is 45 m above water. It falls directly into a small boat moving w...

An object falls from a bridge that is 45 m above water. It falls directly into a small boat moving with constant velocity that is 12 m from the point of impact when the object was released. The speed of the boat is:
A. 2 m/s
B. 3 m/s
C. 5 m/s
D. 4 m/s

Explanation

Solution

Calculate the time taken by the object to fall into a moving boat using a kinematic equation. In this time, the boat has moved a distance of 12 m. Therefore, you can use the relation between distance, velocity and time to determine the speed of the boat.

Formula used:
s=ut+12gt2s = ut + \dfrac{1}{2}g{t^2}
Here, s is the distance covered by the object, u is the initial velocity of the object, g is the acceleration due to gravity and t is the time.

Complete step by step answer:
We have given that the bridge is 45 m above the water. When the ball is dropped such that it directly falls in the boat, the boat has moved a distance of 12 m by the time the ball reaches the boat. Since the boat is moving, we can observe the ball is dropped into the boat even when the boat is not directly at the bottom of the bridge.
We can calculate the time taken by the ball to fall into the moving boat using kinematic equation,
s=ut+12gt2s = ut + \dfrac{1}{2}g{t^2}
Here, s is the distance covered by the ball, u is the initial velocity of the ball, g is the acceleration due to gravity and t is the time.
Since the ball is dropped from that height, we can see the ball has zero initial velocity. Therefore, we can write the above equation as follows,
s=12gt2s = \dfrac{1}{2}g{t^2}
t=2sg\Rightarrow t = \sqrt {\dfrac{{2s}}{g}}
We substitute 45 m for s and 10m/s210\,m/{s^2} for g in the above equation.
t=2(45)10t = \sqrt {\dfrac{{2\left( {45} \right)}}{{10}}}
t=3s\Rightarrow t = 3\,s
Now, we have the relation between distance, velocity and time,
v=dtv = \dfrac{d}{t}
We substitute 12 m for d and 3 s for t in the above equation.
v=12m3sv = \dfrac{{12\,m}}{{3\,s}}
v=4m/s\Rightarrow v = 4\,m/s

So, the correct answer is “Option D”.

Note:
While using the kinematic equation, s=ut+12gt2s = ut + \dfrac{1}{2}g{t^2}, the direction of motion is important. For the object moving upward, you should take the acceleration due to gravity as negative and for the object moving downward, you should take the acceleration due to gravity as positive.