Verified answer

A/

max. height = (V·sinΘ)² / (2g)

h = (24.0m/s * sin42.0º)² / 19.6m/s² = 13.2 m

B/

v = √(V² + 2gh) = √((24.0m/s)² + 2 * 9.8m/s² * 16.0m) = 29.8 m/s

C/

You could either use the vertical data

0 m = 16.0m + 24.0m/s*sin42.0º * t - ½ * 9.8m/s² * t²

to solve the quadratic for t

and then use

x = 24.0m/s * cos42.0º * t

to find the horizontal distance, or simply use the trajectory equation:

y = h + x·tanΘ - g·x² / (2v²·cos²Θ)

and since 1/cos²Θ = sec²Θ = 1 + tan²Θ, we have

y = h + x·tanΘ - (1 + tan²Θ)·g·x² / (2v²)

y = (h - gx²/2v²) + xtanΘ - (gx²/2v²)tan²Θ

where y = height at x-value of interest = 0 m

and h = initial height = 16.0 m

and x = range of interest = ???

and Θ = launch angle = 42.0º

and v = launch velocity = 24.0 m/s

Dropping units for ease,

0 = 16.0 + xtan42.0 - 9.8x² / (2(24.0)²cos²42.0)

0 = 16.0 + 0.9004x - 0.0154x²

I get a negative root (disregard) and a root at x = 72.7 m

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