Answer: from the vertical, one should aim 86.6°
Explanation:
height of the center of object = 7.0 m - 0.05 m = 6.95 m
now let the bullet hits centre at point A height x meters from the ground
also let t be the time taken for the bullet to hit the object
so distance travelled by the target will be
d = h - x = 6.95 - x
now using the equation of motion
d = 1/2gt²
so 1/2gt² = 6.95 - x
x = 6.95 - 1/2gt² .........let this be equ 1
let angle of fire be ∅
so v(cos∅) × t = 100
our velocity v is 1200 ft/sec = 365.76 m/s
365.76(cos∅) × t = 100 ........equ 2
also vertical position of the bullet after t is
y = y₀ + c(sin∅)t - 1/2gt²
y = 1 + 365.76(sin∅)t - 1/2gt² ----- equ 3
After time t. the vertical position x and y are same, else the bullet wouldn't have strike target at centre, so;
x = y
we substitute
equ 1 = equ 3
6.95 - 1/2gt² = 1 + 365.76(sin∅)t - 1/2gt²
6.95 - 1 = 365.76(sin∅)t - 1/2gt² + 1/2gt²
5.95 = 365.76(sin∅)t
t = 5.96 / 365.76(sin∅)
now input the above equ into equ 2
365.76(cos∅) × 5.96/365.76(sin∅) = 100
5.95(cos∅)/sin∅ = 100
tan∅ = 5.95/100 = 0.0595
∅ = 3.40°
therefore from the vertical, one should aim (90° - 3.40°) = 86.6°
