Question

Question 9: A stone is thrown off of a bridge that is 40m above the water. The stone lands 48m from the base of the bridge. The stone was thrown with an initial velocity of 15m/s at 30 degrees above the horizontal. a) What are the initial position coordinates of the stone? b) What are the components of the initial velocity? (Answer: 13m/s, 7.5m/s) c) What is the maximum height the stone reaches above the water? (Answer: 42.87m) d) What are the final position coordinates of the stone? e) What is the final velocity of the stone just before it hits the water? Remember that velocity is a vector. (Answer: 31.8m/s, 66SofE)
Please show work.

Answer 1

here,

9)

the height of bridge , h = 40 m

initial velocity , u = 15 m/s

theta = 30 degree

a)

the initial position is ( 0 , 40 m)

b)

the horizontal component of velocity , ux = u * cos(30)

ux = 13 m/s

the vertical component of velocity , uy = u * cos(30)

uy = 7.5 m/s

c)

the maximum height reached by water , hmax = 40 + uy^2/2g

hmax = 40 + 7.5^2/(2*9.8)

hmax = 42.87 m

d)

the final cooredinates of stone is ( 48 , 0)

e)

vy^2 - uy^2 = 2 * g * h

vy^2 - 7.5^2 = 2 * 9.81 * 40

vy = 29 m/s

the magnitude of velocity , |v| = sqrt(vy^2 + vx^2)

|v| = 31.8 m/s

theta = arctan(29/13) = 66 S of E

final velocity , |v| = sqrt()