Question

A body is thrown upwards with velocity $100m{\text{ }}{s^{ - 1}}$ and it travels $5m$ in the last second of its upward journey if the same body is thrown upward with velocity $200m{\text{ }}{s^{ - 1}}$ what distance will it travel in the last second of its upward journey?

Answer 1

To solve this type of question we must know concepts of kinematics. Here in this question firstly, we have found the final time taken by the body to reach the maximum height and then find the generalize equation of distance (height) as a function of time so then we can calculate the distance travelled in last second and then similarly we have done for $v = 200m{\text{ }}{s^{ - 1}}$ . and hence we got our required solution.

Formula used:
$v = u + at$
$v = ut + \dfrac{1}{2}a{t^2}$
$v$ is the final velocity,
$u$ is the initial velocity,
$a = g$ is the acceleration due to gravity and
$t$ is the time.

Complete step by step answer:
We know that when the body will be at its maximum position the body will have zero velocity,
Therefore, $v = 0$
According to the question the distance travelled by the body in last second is $5m$
And the initial velocity is given i.e., $u = 100m{\text{ }}{s^{ - 1}}$
So, applying equation of motion we get,
$v = u + at$
Now putting all the values in above equation, we will get,
$\Rightarrow 0 = 100 + \left( { - 10(t)} \right)$ Here, $g$ is negative because the body is moving in the opposite direction of force.

$$0 = 100 + \left( { - 10(t)} \right) \\\ \Rightarrow t = \dfrac{{100}}{{10}} \\\ \Rightarrow t = 10s \\\$$

Now, let’s find the generalized equation of distance (height) as a function of time.
$h(t) = ut + \dfrac{1}{2}a{t^2} \\\ \Rightarrow h(t) = 100t + \dfrac{1}{2}10{t^2} \\\ \Rightarrow h(t) = 100t - 5{t^2} \\\$
So, the distance travelled in the last second will be the final distance travelled in total time minus the last second.
i.e., the final time is $10s$ as solved above.
Therefore, the last second will be $9s$ .
Therefore now, distance travelled by the body will be,

$$h(t) = 100t - 5{t^2} \\\ \Rightarrow h(10) - h(9) = \left( {100(10) - 5{{(10)}^2}} \right) - \left( {100(9) - 5{{(9)}^2}} \right) \\\ \Rightarrow h(10) - h(9) = 500 - 495 \\\ \Rightarrow h(10) - h(9) = 5m \\\$$

Now when $u = 200m{\text{ }}{s^{ - 1}}$
Similarly solving the equation,
${t_1} = \dfrac{{200}}{{10}} = 20s$
As same as above,
So, the distance travelled in the last second will be the final distance travelled in total time minus the last second.
i.eThe final time is $20s$ as solved above.
Therefore, the last second will be $19s$ .
Therefore now, distance travelled by the body will be,

$$h(t) = 200t - 5{t^2} \\\ \Rightarrow h(20) - h(19) = \left( {200(20) - 5{{(20)}^2}} \right) - \left( {200(19) - 5{{(19)}^2}} \right) \\\ \Rightarrow h(20) - h(19) = 2000 - 1995 \\\ \Rightarrow h(20) - h(19) = 5m \\\$$

Therefore, the body traveling in the last second of its upward journey is $5m$ .

Note: Note that the generalized equation of distance (height) as a function of time helps us easily calculate the height (distance) at whatever time we want to find. The secret here is that the acceleration owing to gravity is treated as negative rather than positive upwards, i.e., the y direction is deemed positive, but g will act downwards, or in the negative y direction, thus it is taken as negative. The equation of motion is valid if and only if the acceleration of the body remained constant throughout its journey, and since it was constant as g, the laws of motion apply.