Question

A rock is released from rest at a certain height and falls freely, reaching an impact speed of \[4\dfrac{m}{s}\] at the floor. Next, the rock is thrown down with an initial speed of \[3\dfrac{m}{s}\] from the same height. What is its speed on the floor?
     \[\]
A. \[14\dfrac{m}{s}\]
B. \[25\dfrac{m}{s}\]
C. \[8\dfrac{m}{s}\]
D. \[5\dfrac{m}{s}\]

Answer 1

Hint: Here we used the equations of motion to solve this question. We also know, If an object is dropped the initial velocity of the object is zero. In this question by substituting the initial velocity and the final velocity given in the question we will find the total height and then from the second part of the question we will find the final velocity of rock \[2\].

Complete step by step solution:
According to the question the rock falls freely from a certain height.
For, the \[{1^{st}}\] rock, we applied the third equations of motion -
 \[{v^2} = {u^2} + 2as\]
Here given, \[u = 0\]
 \[v = 4\dfrac{m}{s}\]
now putting the values in above equation-
\[{(4)^2} = {(0)^2} + 2(10)s\] [assuming \[g = 10\]]
\[ \Rightarrow 16 = 0 + 20s\]
\[ \Rightarrow 16 = 20s\]
\[ \Rightarrow s = 0.8\]
For \[{2^{^{nd}}}\] rock we also apply the\[{3^{rd}}\]equation of motion-
But this time \[u = 3\dfrac{m}{s}\] and \[s\] is remaining same because the height also remain same,
\[
   \Rightarrow {v^2} = {(3)^2} + 2 \times 10 \times 0.8 \\
   \Rightarrow {v^2} = 9 + 16 \\
   \Rightarrow {v^2} = 25 \\
   \Rightarrow v = 5 \\
 \]
Therefore, the option D) \[5\dfrac{m}{s}\] is the right answer.

Note:
A body is at rest which implies that it isn’t moving, simply implies that it is being depicted reading a casing of reference that is moving along with the body. For instance, a body on the outside of the Earth may give off an impression of rest, yet that is simply because the observer is on the Earth’s surface.