Question

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

Answer 1

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.