- #1
Daniel Dubois
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Homework Statement
Golf ball struck with a velocity of 66 ft/s.
Part a): determine the speed at which the ball hits point B.
Part b): determine the time of flight from A to B.
Homework Equations
For Part b):
(Vo)x = Vcos(θ)
(Vo)y = Vsin(θ)
D = Do + (Vo)x * t
Y = Yo + (Vo)y * t - (1/2) * g * t^2
For Part a):
Vy^2 = (Vo)y^2 - 2 * g * (Y-Yo)
Vx^2 = (Vo)x^2 - 2 * g * (Y - Yo)
V^2 = [(V)x^2 + (V)y^2]
V = [(Vo)x + (Vo)y] / 2
The Attempt at a Solution
If there are more effective and less complicated ways of solving the asked questions please let me know.
I solved for time first (part b).
-- given an initial velocity Vo in the problem statement of 66 ft/s
-- from diagram angle between horizontal and initial direction of the ball 10° + 45° = 55°
(Vo)x = Vcos(θ) -------> (Vo)x = 66cos(55) -----> (Vo)x = 37.89 ft/s
(Vo)y = Vsin(θ) -------> (Vo)y = 66sin(55) -----> (Vo)y = 54.06 ft/s
-- solving for distance d
S = So + (Vo)x * t -------> d * cos(10) = 0 + 37.89 * t ----> d = 38.44 * t
-- solving for final height of ball (at point B).
Y = Yo + (Vo)y * t - (1/2) * g * t^2 -----> d * sin(10) = 0 + 54.06 * t - 16.1 * t^2 -----> 6.68 * t = 54.06 * t - 16.1 * t^2
-- solving for time t
t = 2.94s This answer was taken as correct in online program.
-- Plugging in t value in d
d = 38.44 ft
-- Plugging in d value in Y = d * sin(10)
Y = 19.62
For my attempt at part a) final speed of the projectile at point B I tried two different formulas, but with no success.
Vy^2 = (Vo)y^2 - 2 * g * (Y-Yo) -----> Vy^2 = 54.06^2 - 2 * 32.2 * (19.62 - 0) -----> Vy = 40.73 ft/s
Vx^2 = (Vo)x^2 - 2 * g * (S-So) -----> Vx^2 = 37.86^2 - 0 -----> Vx = 37.86 ft/s
V^2 = [(V)x^2 + (V)y^2]
V = [(Vo)x + (Vo)y] / 2
The fourth equation doesn't make much sense to me, and with the correct values inserted answer resulted incorrect. The third gives me a really high number which also resulted incorrect.