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Ryan throws a tennis ball straight up into the air. The ball reaches its maximum height at 2 seconds. The approximate height of the ball x seconds after being thrown is shown in the table.

y = –17(x)(x – 4)
y = –16(x)(x – 4)
y = –16(x – 2)^2 + 68
y = –17(x – 2)^2 + 68

Respuesta :

MissPhiladelphia
The equation for a ball thrown straight up is:
y = v₀t - 0.5gt²

We don't know the initial velocity of the ball. We use the given condition from the problem that the maximum height was achieve after t = 2
y = 2v₀ - 19.62
v₀ = 0.5y + 9.81

Substituting, factoring, and replacing t with x, the result is:
y = –16(x – 2)^2 + 68

Answer:

its option C on edg.

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