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Question: A body is projected vertically up to \(t=0\) with a velocity of \(98 ms^{-1}\). Another body is proj...

A body is projected vertically up to t=0t=0 with a velocity of 98ms−198 ms^{-1}. Another body is projected from the same point with the same velocity after 4 seconds. Both bodies will meet after:
A. 6s
B. 8s
C. 10s
D. 12s

Explanation

Solution

A good starting point would be to consider a relevant kinematic equation of motion that you can substitute into this scenario. In other words, you can assume an unknown height where the two bodies meet and use the equation relating height(distance), initial velocity, time and g with respect to both the bodies, accounting for the time difference between their launch.

Formula used:
Kinematic equation of motion: h=ut−12gt2h = ut - \dfrac{1}{2}gt^2 where h is height a body travels, u is the initial projection velocity, t is the time taken, g is the acceleration due to gravity. The negative sign is indicative of a body moving against gravity.

Complete step by step answer:
We have two bodies that are projected vertically up with the same velocity but at 4 seconds apart.
Let us assume that they meet at a time t and at a height h. Using a kinematic equation of motion:
h=ut−12gt2h = ut - \dfrac{1}{2}gt^2
Body 1: h=98  t−12×9.8×t2h = 98\;t - \dfrac{1}{2} \times 9.8 \times t^2
Body 2: h=98(t−4)−12×9.8×(t−4)2h = 98\left(t-4\right) - \dfrac{1}{2} \times 9.8 \times \left(t-4\right)^2
Equating the above two expressions we get:
98  t−12×9.8×t2=98(t−4)−12×9.8×(t−4)2⇒98×4+9.82(16−8t)=0⇒392=−  4.9(16−8t)98\;t - \dfrac{1}{2} \times 9.8 \times t^2 = 98\left(t-4\right) - \dfrac{1}{2} \times 9.8 \times \left(t-4\right)^2 \Rightarrow 98\times 4 +\dfrac{9.8}{2}\left(16-8t\right) = 0 \Rightarrow 392 = -\;4.9\left(16-8t\right)
⇒3924.9=(8t−16)⇒80=(8t−16)⇒96=8t⇒t=12  s\Rightarrow \dfrac{392}{4.9} = \left(8t-16\right) \Rightarrow 80 = \left(8t-16\right) \Rightarrow 96 =8t \Rightarrow t= 12\;s

So, the correct answer is “Option D”.

Note:
The only difference between the projected bodies here is the time at which they are thrown vertically up. However, one can expect variations in this type of a question where the initial velocity or time or both may be subject to change.
Notice that the equation that we’ve used above is just a tweaked form of the equation of translational motion where we have just replaced the distance travelled with height reached, and the acceleration is due to gravity that is directed downwards, and since the particle is moving in a direction opposite to that action of gravity g gains a negative sign.