Need help finding final speed of a projectile

AI Thread Summary
The discussion focuses on calculating the final speed of a golf ball projectile and its time of flight. The initial velocity of the ball is 66 ft/s, with a launch angle of 55 degrees. The user successfully calculated the time of flight as 2.94 seconds and the horizontal distance as 38.44 ft. However, they encountered difficulties in determining the final speed at point B, trying multiple formulas without success. Suggestions were made to consider the incline's frame of reference and to use components of gravitational acceleration for a more effective solution.
Daniel Dubois
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Homework Statement


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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.


 
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Daniel Dubois said:

The Attempt at a Solution


If there are more effective and less complicated ways of solving the asked questions please let me know.

For b). Think about the horizontal motion and remember Mr Pythagoras!
 
You could write the equation for vertical displacement from the incline's frame of reference(i.e. angle of projection=θ=45°). For that, you'll need components of gravitational acceleration along and normal to the incline. Equation for vertical displacement will be a quadratic equation in t. From incline's frame of reference, vertical displacement will be 0 when the ball reaches point B. Just solve it to get the time of flight.
 
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