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A shot is fired from apoint of adistance of 200m from the foot of the tower 100 high so that it just passes over it the direction of shoot is

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by gshreya681

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by gshreya681

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now vertical component of the velocity = vsinθ

as the shot just passes over the tower, with the help of the vertical component of the velocity it will reach the height of the tower 100m & at the top of the tower vsinθ = 0

now using the formula "v² = u² - 2gh" we get

so here v = 0, u = vsinθ at the ground, h = height of the tower

so vsinθ at ground = √2gh = √(2×9.8×100) = 44.27 m/s

time required to reach the top of the tower = 44.27/9.8 = 4.5 sec ( using t = u/g )

again horizontal component of the velocity = vcosθ

with the help of the horizontal component of the velocity vcosθ the shot just covers a distance of 200m in the time 4.5 sec .

so vcosθ = 200/4.5 = 44.44 m/s

now vcosθ = 44.44m/s

vsinθ = 44.27m/s

so tanθ = 44.27/44.44 = 0.996 ≈ 1

so θ = tan⁻¹1 = 45⁰

so the shot should be projected by making an angle 45⁰ with the horizontal.