Need help finding final speed of a projectile

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SUMMARY

The discussion focuses on calculating the final speed and time of flight of a golf ball struck at an initial velocity of 66 ft/s. The projectile motion equations used include horizontal and vertical components derived from the angle of projection, specifically 55°. The time of flight was successfully calculated as 2.94 seconds, leading to a horizontal distance of 38.44 ft. However, the final speed at point B was incorrectly computed using various formulas, indicating a need for clarification on the correct application of projectile motion equations.

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  • Understanding of projectile motion principles
  • Familiarity with trigonometric functions (sine and cosine)
  • Knowledge of kinematic equations for motion
  • Basic grasp of gravitational acceleration (g = 32.2 ft/s²)
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  • Review the derivation and application of projectile motion equations
  • Learn about the components of gravitational acceleration in inclined frames
  • Study the use of quadratic equations in solving motion problems
  • Explore the concept of vector addition in determining resultant velocities
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Students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators seeking to clarify concepts related to kinematics.

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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