The following parametric equations model the flight of the baseball

Question

A baseball is hit from a height of 3 feet at a 60° angle, with initial velocity 64 feet per second. The following parametric equations model the flight of the baseball:

𝑥 = 32t

𝑦 = −16𝑡 2 + 32√3 𝑡 + 3

a. How much time does it take until the ball hits the ground? Round to the nearest tenth, and include units in your answer.

b. What horizontal distance did the ball travel before it hit the ground? Round to the nearest tenth, and include units in your answer.

Answer 1

The ball hitting the ground would be when it reaches a height of 0

0 = −16 t² + 32 t + 3

t > 0 because any time less than 0 (a "negative time") would not make sense in the context of throwing the ball and progressing forward in time.

The following solution for t is the application of the quadratic formula with:

a = -16, b = +32 √3, & c = +3

t = [ -32 √3 - √( ( -32 √3 )² - 4 ( -16 ) ( 3 ) ) ] / [ 2 (-16) ]

t = √3 + √( 3264 ) / 32

∴ t = √3 + √( 51 ) / 4 ≈ 3.5 s

x ( √3 + √( 51 ) / 4 ) = 32 ( √3 + √( 51 ) / 4 )

∴ x ( √3 + √( 51 ) / 4 ) = 32 √3 + 8 √( 51 ) ≈ 112.6 ft

I hope this helps! Message me in the comments with any questions, comments, or concerns!

Answer 2

The ball is at ground level when y = 0.

0 = −16𝑡² + 32√3 𝑡 + 3

Using the quadratic formula:

t = [-b ± √(b² - 4ac)]/(2a)

t = [-32√3 ± √((32√3)² - 4(-16)(3))]/(2(-16))

t = [-32√3 ± √(3072 + 192)]/(-32)

t = [-32√3 ± √3264]/(-32)

t = [-32√3 ± 8√51]/(-32)

t = √3 ± √51/4

The "minus" answer is not valid therefore the ball hits the ground in √3 + √51/4 seconds.

The horizontal distance it travels is calculated from the "x" equation: x = 32(√3 + √51/4) = 32√3 + 8√51 feet