Solving for Minimum Angle for Home Run

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SUMMARY

The discussion focuses on calculating the minimum launch angle (theta) required for a baseball to clear a 10-foot high fence located 400 feet from home plate. The ball is hit at a speed of 147.67 ft/s from a height of 3 feet. The projectile's path is modeled using parametric equations, where the horizontal motion is defined by x = v0cos(theta)t and the vertical motion by y = 3 + (v0sin(theta)t - 15t^2). The solution involves substituting values into these equations and transforming them to find theta using the relationship between secant and tangent functions.

PREREQUISITES
  • Understanding of projectile motion equations
  • Familiarity with trigonometric identities, specifically secant and tangent
  • Basic algebra for solving quadratic equations
  • Knowledge of parametric equations in physics
NEXT STEPS
  • Study the derivation of projectile motion equations in physics
  • Learn how to apply trigonometric identities in physics problems
  • Practice solving quadratic equations in the context of projectile motion
  • Explore the effects of initial velocity and angle on projectile trajectories
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the mathematics behind projectile motion and its applications in sports, particularly baseball.

jazz20
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Homework Statement


the center field fence in a ball park is 10ft high and 400 feet from home plate. the ball is hit 3 ft above ground. it leaves the bat at an angle of theta degrees with the horizontal speed of 147.67ft/second. find the minimum angle at which the ball must leave the bat in order for hit to be a home run.
the path of the projectile is modeled by the parametric equations:
x=v0cos(theta)t
y=3+(v0sin(theta)t-15t^2

Homework Equations





The Attempt at a Solution


substitute 400 in for x and 147.67 for v0to solve for t
t=400/(146.67cos(theta)
then i took t and put it into the equation for y
y=3+(147.67sin(theta)(400/(146.67cos(theta))-16(400/(146.67cos(theta))^2


Did I do this correctly? If so, can some one help me with the math to solve for theta?
 
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jazz20 said:
1.

The Attempt at a Solution


substitute 400 in for x and 147.67 for v0to solve for t
t=400/(146.67cos(theta)
then i took t and put it into the equation for y
y=3+(147.67sin(theta)(400/(146.67cos(theta))-16(400/(146.67cos(theta))^2


You can rewrite the equation as
y = 3 + 400*tanθ - 16*400^2*sec^2θ/146.67^2 [ 1/cosθ = secθ]
Put sec^2θ = 1 + tan^2θ,and solve the quadratic for tanθ.
 

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