William and Stephanie are playing a rather dangerous game of catch on the rooftops of skyscrapers in downtown Chicago. William is preparing to throw a tennis ball off the roof of the John Hancock Center, 1130 feet above street level. Stephanie is waiting to catch the ball on the rooftop of the neighboring Water Tower Place, 830 feet above street level. William throws the ball, which leaves his hand with an upward velocity of 46 feet per second, and travels in a parabolic path. Stephanie catches the ball exactly 6.0 seconds after it is thrown. Create a quadratic function describing the height of the ball above the street level with respect to time.

Question

Alyssa Michael

Mathematics

2 November, 18:09

William and Stephanie are playing a rather dangerous game of catch on the rooftops of skyscrapers in downtown Chicago. William is preparing to throw a tennis ball off the roof of the John Hancock Center, 1130 feet above street level. Stephanie is waiting to catch the ball on the rooftop of the neighboring Water Tower Place, 830 feet above street level. William throws the ball, which leaves his hand with an upward velocity of 46 feet per second, and travels in a parabolic path. Stephanie catches the ball exactly 6.0 seconds after it is thrown. Create a quadratic function describing the height of the ball above the street level with respect to time.

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Courtney Humphrey · Answer 1

The question involves the concept & equations associated with projectile motion.

Given:

y₁ = 1130 ft

v₁ = + 46 ft/s (note positive sign indicates upwards direction)

t = 6.0 s

g = acceleration due to gravity (assumed constant for simplicity) = - 32.2 ft/s²

Of the possible equations of motion, the one we'll find useful is:

y₂ = y₁ + v₁t + 1/2gt²

We can just plug and chug to define the equation of motion:

y = (1130 ft) + (46 ft/s) t + 1/2 (-32.2 ft/s²) t²

(note: if you were to calculate y using t = 6.0 s, you'd find that y = 826.4 ft, instead of 830 ft exactly because of some rounding of g and/or the initial velocity)