Question

a golfer hits a golf ball an initial velocity of 54m/s [42 degrees above the horizontal]. the ball lands on an elevated green that is 14m higher that the level from which the golfer hit the ball.

a) how far away horizontally from the golfer does the ball land?

b) what is the velocity of the ball when it lands?

please show all the steps

Answer 1

Vo = 54m/s[42o]
Xo = 54*Cos42 = 40.1 m/s.
Yo = 54*sin42 = 36.1 m/s.

a. Y = Yo + g*Tr = 0
Tr = -Yo/g = -36.1/-9.8 = 3.69 s. = Rise
time.

Y^2 = Yo^2 + 2g*h = 0
h = -(Yo^2)/2g = (36.1^2)/-19.6=66.5 m.

0.5g*t^2 = 66.5-14 = 52.5
4.9t^2 = 52.5
t^2 = 10.71
t = 3.27 s. = Fall time(Tf).

Dx = Xo * (Tr+Tf) = 40.1 * (3.69+3.27) =
279.1 m. = Hor. distance.

b. Y^2 = Yo^2 + 2g*h
Y = Ver. component of final velocity.
Yo = 0
g = +9.8 m/s^2
h = 66.5-14 = 52.5 m.
Solve for Y.

V = sqrt(Xo^2+Y^2) m/s