Question

3. An assassin who climbed to the top of the catwalk balcony of a mansion, to get a
good view of his target who was 6.1 ft tall. The assassin pulled the trigger when
the target is 294 ft away, discharging the bullet at a 63° angle to the level of the
victim's head. Find the height that the assassin shot the target from.

Answer 1

The height of the mansion that the assassin shot the target from is determined as 1,164.9 ft.

Time of motion of the bullet

The time of motion of the bullet is calculated as follows;

-h = vsinθ(t)  - ¹/₂gt²

-6.1 = (vt) sin63  - ¹/₂(32)t²

-6.1 = 0.89vt - 16t² ---(1)

From horizontal motion

X = vcosθ t

294 = (vt)cos63

294 = 0.454vt

vt = 294/0.454

vt = 647.577 ---(2)

solve (1) and (2) together;

-6.1 = 0.89(647.577) - 16t²

16t² = 576.344 + 6.1

16t² =  582.44

t² = 582.44/16

t² = 36.4

t = √36.4

t = 6.03 s

Initial velocity

v = 647.577/t

v = 647.577/6.03

v = 107.4 ft/s

Height of the mansion

h = vt + ¹/₂gt²

h = (107.4 x sin63 x 6.03) + ¹/₂(32)(6.03)²

h = 1,158.8 ft

H = 1,158.8 ft + 6.1 ft = 1,164.9 ft

Thus, the height of the mansion that the assassin shot the target from is determined as 1,164.9 ft.

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