height velocity formula
2009
If you throw a ball, which reaches its apex at 100 feet to 1.5 seconds, how to find the initial velocity and height?
Using the formula h (t) = -1/2gt ^ 2 + VT + s, and v is the initial velocity, the "s" is the original height, "t" is time, and g is 32 m / s / sec. In addition, after 4 seconds to reach the ground, and you are in a building.
Here is the function of its position original ball: h (t) = -1/2gt ^ 2 + VT + s This is the first derivative of this function: h (t) = V (t) =- 32t + v After the derivative is now the function of speed here, I'm just demonstrating that in 1.5 seconds, speed is zero. V (1.5) = 0 m / s V (1.5) = 0 = -32 (1.5) + vv = ————–> 32 * 1.5 V = 48 m / s Since it hits the ground after 4 seconds, h (4) = 0 = – 1 / 2 (32 ) (4 ^ 2) + 4 (48) S + s = 1 / 2 (32) (4 ^ 2) – 4 (48) s = 64 m
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