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`        plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz solve question no. 19`
one year ago

```							Here, u = 150 ms-1And, Taking g = 10 m-2 (approximately) Let 'θ' be the angle at which the machine gun should fire in order to cover a maximum distance. Then , the horizontal component of velocity = 150 cos θ And, the vertical component of velocity = 150 sin θ If 'T' is the time of flight, then , Horizontal range, R = (150 cos θ ) x T . The gun is mounted at the top of a tower 100 meters high . Let us regard the positive direction of the position-axis as to be along the line from the top of tower in downward direction. For motion along vertical :Initial Velocity = -150 sin θ; Distance covered = + 100 m And, acceleration = + 10 ms-2In time 'T' , the machine gun shot will reach maximum height and then reach the ground. Now, S = ut + 1/2 at2Therefore, +100 = ( -150 sin θ ) T + 1/2 x T2Or, T2 - ( 30 sin θ )T - 20 = 0 Therefore, T = - ( - 30 sin θ ) ± { ( - 30 sin θ)2 - 4 x 1 x (-20) }1/2/2 = 30 sin θ ± (900 sin2 θ + 80 )1/2/2 Or, T = 15 sinθ ± (225 sin2θ + 20)1/2Now, range will be maximum, if time of flight is maximum . Therefore, choosing positive sign ,we have T = 15 sin θ + ( 225 sin2θ+ 20 )1/2Hence, horizontal range covered, R = 150 cos θ {15 sin θ+ ( 225 sin2θ + 20 )}1/2The horizontal range is maximum, when θ = 45o. But in the present case , the machine gun is mounted at height of 100 m . Therefore, R will not be maximum for θ = 45o.It will be maximum for some value of θ close to 45o. If we calculate values of R by setting θ = 43o, 43.5o, 44o, 45o, 46o and 47o, the values of R come out to be 2347 m, 2347.7 m, 2348 m, 2346 m, 2341 m and 2334 m respectively. Thus R is maximum for value of θ some where between 43.5o and 44o. Therefore , the mean value of θ = (43.5o + 44o)/2 = 43.75o. The gun should be inclined at 43.75o to cover a maximum range of firing on the ground below.
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one year ago
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