Question

While sitting on a tree branch 20m above the ground, you drop a chestnut. When the chestnut has fallen 5m, you throw a second chestnut straight down. What initial speed must you give the second chestnut if they are both to reach the ground at the same time? (\[g = 10\,{\text{m/}}{{\text{s}}^2}\])
A. \[5\,{\text{m}}{{\text{s}}^{ - 1}}\]
B. \[10\,{\text{m}}{{\text{s}}^{ - 1}}\]
C. \[15\,{\text{m}}{{\text{s}}^{ - 1}}\]
D. None of these

Answer 1

Hint: Calculate the time taken by the first chestnut to reach the ground. Then calculate the time taken by the first chestnut to cover 5 m distance. The time taken by the second chestnut to reach the ground is equal to the time taken by the first chestnut to reach the ground from 5 m initial distance.

Formula used:
\[s = ut + \dfrac{1}{2}a{t^2}\]
Here, s is the distance, u is the initial velocity, a is the acceleration and t is the time.

Complete step by step answer:
We have given that the total distance that to be travelled by both the chestnuts is \[h = 20\,{\text{m}}\]. Since we have dropped the first chestnut from the tree branch, the initial velocity of the first chestnut is zero. The second chestnut is given some initial velocity to reach the ground in the same time as the first chestnut does in the last 15 m.

We have to determine the time taken by the first chestnut to reach the ground. Let’s use the kinematic equation to express the total distance travelled by the first chestnut as follows,
\[h = ut + \dfrac{1}{2}g{t^2}\]

Here, u is the initial velocity, g is the acceleration due to gravity and t is the time taken by the first chestnut to reach the ground.

Since the initial velocity is zero, the above equation becomes,
\[h = \dfrac{1}{2}g{t^2}\]
\[ \Rightarrow t = \sqrt {\dfrac{{2h}}{g}} \]

Substituting \[h = 20\,{\text{m}}\] and \[g = 10\,{\text{m/}}{{\text{s}}^2}\] in the above equation, we get,
\[t = \sqrt {\dfrac{{2\left( {20} \right)}}{{10}}} \]
\[ \Rightarrow t = 2\,{\text{s}}\]

Now, the time for which the first chestnut was in the air when it travelled a distance 5 m is,
\[t' = \sqrt {\dfrac{{2h'}}{g}} \]

Substituting \[h' = 5\,{\text{m}}\] and \[g = 10\,{\text{m/}}{{\text{s}}^2}\] in the above equation, we get,
\[t' = \sqrt {\dfrac{{2\left( 5 \right)}}{{10}}} \]
\[ \Rightarrow t' = 1\,{\text{s}}\]

Therefore, the first chestnut takes time to travel the rest 15 m distance is,
\[t = 2\,{\text{s}} - 1\,{\text{s}}\]
\[ \Rightarrow t = 1\,{\text{s}}\]

The second chestnut should be thrown with initial velocity such that it reach the ground in \[t = 1\,{\text{s}}\].

Using the same kinematic equation as above, we get,
\[h = ut + \dfrac{1}{2}g{t^2}\]

Substituting \[h = 20\,{\text{m}}\], \[t = 1\,{\text{s}}\] and \[g = 10\,{\text{m/}}{{\text{s}}^2}\] in the above equation, we get,
\[20 = u\left( 1 \right) + \dfrac{1}{2}\left( {10} \right){\left( 1 \right)^2}\]
\[ \Rightarrow 20 = 2u + 5\]
\[ \Rightarrow u = \dfrac{{15}}{2}\]
\[ \therefore u = 7.5\,{\text{m/s}}\]
Therefore, the second chestnut should be thrown with initial velocity equal to 7.5 m/s.

So, the correct answer is option D.

Note: For the downward motion of the body, choose the acceleration due to gravity and distance travelled as positive since their direction is downwards. The acceleration due to gravity will be the same in both the cases since it is the constant quantity. One should not use the fact that the second chestnut should be given the initial velocity so that it reaches the ground at the same time as the first chestnut does. The time for the first chestnut to travel 15 m distance should be the same as the time taken by the second chestnut to reach the ground from its initial height.