Answered

IDNLearn.com is your go-to resource for finding answers to any question you have. Ask anything and get well-informed, reliable answers from our knowledgeable community members.

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.

Sagot :

Onyebuchinnajiidn

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.

Learn more about time of motion here: https://brainly.com/question/2364404
#SPJ1

We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Your search for answers ends at IDNLearn.com. Thank you for visiting, and we hope to assist you again soon.