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The speed with which a bullent can be...

The speed with which a bullent can be fired is `150 ms^(-1)`. Calculate the greatest distance to which it can be projected and also the maximum height to which it would rise .

Text Solution

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The greatest horizontal range is achieved at an angle of projectionn of `45^(@)`
Components of initial velocity `=150cos45^(@)=106.06ms^(-1)`
Now, if T is the time of flight, then considering the vertical motionof the bullet,
`u=106.06 ms^(-1),a=-9.8ms^(-1),s=0,t=T`
using `s=ut+1/2at^(2)`
we get `0.106.06T-1/2 9.8T^(2)impliesT=(106.06xx2)/9.8=21.64` sec
`:.` Maximum Horizontal range = horizontal component of velocity `xx` total time of flight.
`=106.06xx21.64=2295.14m`
Again if `H_("max")` be the maximum height of when the bullet rises, then
`u=106.06ms^(-1),a=-9.8ms^(-1)`
`v=0,s=H_("max")=(u^(2))/(2)impliesH_("max")=((106.06)^(2))/19.6=573.91m`
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