Q13 / 72 · NTA Abhyas (JEE Main) 2020 · Physics · Motion In Two Dimensions

A shell bursts on contact with the ground and pieces from it fly in all directions with a velocity equal to 60 m s-1. If it is known that a man who is 180 m away is in danger for a time interval of x seconds, then what is the value of x? [Neglect the height of the man]

Correct answer: 72

Solution

Given, u=60 m s-1, R=180 m 

∴  R=u2gsin⁡2θ

180=60210sin⁡2θ

⇒ sin⁡2θ=12

⇒ θ=15o

As there are always two directions of projections θ and 90o-θ for the given range, hence two directions of projection, i.e. 15o and 90o-15o, i.e. 15o and 75o. Let T1 and T2 be the times of flight in the two cases, where

T1=2usin⁡15og and T2=2usin⁡75og

The man is in danger for a time

T2-T1=2ugsin⁡75o-sin⁡15o

=2ug2cos⁡45osin⁡30o

using identity, sin⁡C-sin⁡D=2 cos⁡C+D2sin⁡C-D2

=2×6010×2×12×12

=72 s

Hence, the correct answer is 72.

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