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A projectile is launched from a tower 64 feet above the ground into the air with an initial velocity of 48 feet per second. How many seconds is the projectile in the air before it hits the ground?

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nimrah3

Answer:

I hope this helps you

Step-by-step explanation:

To define the time of flight equation, we should split the formulas into two cases:

Launching projectile from the ground (initial height = 0)

t = 2 * V₀ * sin(α) / g.

Launching projectile from some height (so initial height > 0)

t = [V₀ * sin(α) + √((V₀ * sin(α))² + 2 * g * h)] / g.

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