Question

During a baseball practice session, the cutoff man A executes a throw to the third baseman B. If the initial speed of the baseball is v₀ = 123 ft/sec, what angle θ is best if the ball is to arrive at third base at essentially ground level?

6.2 A 153

Answer 1

In horizontal,

horizontal velocity, v0x = 123 cos(theta) { that is constant throughout }

time of flight = Dx / V0x = 153 / (123 cos(theta))

In vertical,

y = - 6.2 ft

ay = - 32.174 ft/s^2

v0y = 123 sin(theta)

y = v0y t + ay t^2 / 2

- 6.2 = 123 sin(theta) - 32.174 (153 / 123 cos(theta))^2 / 2

- 6.2 = 123 sin(theta) - 24.89/ (1 - sin^2(theta))

-6.2 + 6.2 sin^2 (theta) = 123 sin(theta) - 123 sin^3 (theta) - 24.89

{ x = sin(theta)}

123 x^3 + 6.2x - 123x + 18.69 = 0

x = 0.165, 0.882

sin (theta) = 0.165 , 0.882

theta = 9.50 deg or 62 deg