A cannon and a target are $5.1 k m$ apart located on the same level. Find the time to hit the target. Speed of cannon is $240 m s^{- 1}$ . Assume no air drag.

24.6 s

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25.9 s

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23 s

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21 s

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Solution

The correct option is B $24.6 s$
Using $R = \frac{u^{2} sin 2 θ}{g}$ or $sin 2 θ = \frac{R g}{u^{2}} = \frac{5100 \times 10}{(240)^{2}} = 0.88$
or $2 θ = 61^{o} 42^{'} \Rightarrow θ = 30^{o} 51^{'}$
Time for hitting target is $T = \frac{2 u sin θ}{g} = \frac{2 \times 240 sin 30^{o} 51^{'}}{10} = 24.6 s$

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A cannon and a target are 5.1km apart located on the same level. Find the time to hit the target. Speed of cannon is 240ms−1. Assume no air drag.

The correct option is B 24.6sUsing R=u2sin2θg or sin2θ=Rgu2=5100×10(240)2=0.88or 2θ=61o42′⇒θ=30o51′Time for hitting target is T=2usinθg=2×240sin30o51′10=24.6s

A cannon and a target are $5.1 k m$ apart located on the same level. Find the time to hit the target. Speed of cannon is $240 m s^{- 1}$ . Assume no air drag.