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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



Let, the shot is fired with velocity v making an angle θ with the horizontal direction 
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.
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