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You are on top of a building that is 75.0 m tall. You toss a ball straight up with an initial velocity of 33.8 m/s. How high does the ball travel? It goes up and then falls down to the ground below. How much time is it in the air?
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@jim_thompson5910
the general equation we'll use is y = -4.9x^2+v*x + h where... initial height = h = 75 initial velocity = v = 33.8 x = time y = height at time x with me so far?
yes
So y = -4.9x^2+v*x + h turns into y = -4.9x^2+33.8*x + 75
are you able to find the vertex of `y = -4.9x^2+33.8*x + 75` ?
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Could I use the formula x=v^2/2a ?
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