Question

A body is thrown upwards with velocity 40$m{{s}^{-1}}$ and it covers 5m in the last second of its upward journey. If the same body is thrown upwards with velocity 80$m{{s}^{-1}}$, what distance will it travel in the last second of the upward journey? (Take g=10$m{{s}^{-2}}$)
A. 5m
B. 10m
C. 15m
D. 20m

Answer 1

Hint: It is given that the initial velocity of the body in the second case is 80 $m{{s}^{-1}}$. We can find the time taken to reach the highest point and the distance that the body covers in this time by using the kinematic equations. Then we can calculate the distance that it covers before the last second. Then subtract the two distances, to find the distance covered in the last second.

Formula used:
$v=u+at$
$2as={{v}^{2}}-{{u}^{2}}$
$s=ut+\dfrac{1}{2}a{{t}^{2}}$

Complete answer:
It is given that the body is thrown upwards with a velocity of 80$m{{s}^{-1}}$ and t we are supposed to find the distance that it covers in the last second of its upwards journey.
For this, let us first find the time taken by the body to reach its maximum height. Then we can calculate the total distance that it covers while going to the highest point.
Let the time taken to reach the maximum height be t. To find the time t we will use the kinematic equation $v=u+at$ ….. (i),
v and u are the final initial velocities of the body and a is the acceleration of the body.
In this case, u = 80$m{{s}^{-1}}$. The body will stop at the maximum height. Therefore, v = 0.
Since the body is accelerated by the gravity in the downwards direction, a = -g = -10$m{{s}^{-2}}$.
Substitute the values of v, u and a in equation (i).
$\Rightarrow 0=80+(-10)t$
$\Rightarrow t=8s$.
Now, let us find the distance that body covers in this time. Let that distance be s.
 To find s, we will use the kinematic equation $2as={{v}^{2}}-{{u}^{2}}$.
$\Rightarrow 2(-10)s={{0}^{2}}-{{(80)}^{2}}$
$\Rightarrow 20s=6400$
$\Rightarrow s=320m$.
Now we can find the distance that the body covers before the last second. Let this distance be s’.
To cover the distance of s’, the body took a time of (t-1)=7s.
Use the kinematic equation $s=ut+\dfrac{1}{2}a{{t}^{2}}$.
 $\Rightarrow s'=(80)(7)+\dfrac{1}{2}(-10){{(7)}^{2}}$
$\Rightarrow s'=(80)(7)-(5)(49)$
$\Rightarrow s'=560-245=315m$.
Therefore, in the last second of the inward journey, the body will travel will travel a distance equal to s - s’ = 320 – 315 = 5m

Hence, the correct option is A.

Note:
We can solve the given question in a different way.
For this, we will use the kinematic equation $v=u+at$ for both motions of the body.
Let's consider the last second in both the motions.
In the first motion, where the body is thrown with velocity of 40$m{{s}^{-1}}$, the velocity after the last second is zero, i.e. v=0. The acceleration is a=-10$m{{s}^{-2}}$ and t =1s.
Let us find the velocity just before the last second, i.e. ${{u}_{1}}$.
$\Rightarrow 0={{u}_{1}}+(-10)(1)$
$\Rightarrow {{u}_{1}}=10m{{s}^{-1}}$.
Similarly, will we find that the velocity of the body just before, the last second is the same. i.e. 10 $m{{s}^{-1}}$.
Now, from the equation $s=ut+\dfrac{1}{2}a{{t}^{2}}$, we know that in the last second the body will cover a distance of $s=u+\dfrac{1}{2}a$.
Since the value of u and a are the same for this time in both cases, the value of s will be the same.
Therefore, in both cases the body covers the same distance, in the last second.