Acoin is tossed vertically upward from the edge of the top of a tall building. the coin reaches its maximum height above the top of the building 1.29 s after being launched. after barely missing the edge of the building as it falls downward, it strikes the ground 6.30 s after it was launched. how tall is the building? we assume an answer in meters

115 m

Explanation:

The coin is in free fall, subject only to the acceleration of gravity. We set up a frame of reference with the 0 at ground and the X axis pointing upwards. We use the equation for position under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

In this case X0 will be the height of the building as it is the starting point of the coin.

We have one equation with 2 unknowns (V0 and X0), we need another equation.

The equation for speed under constant acceleration is:

V(t) = V0 + a * t

We know that at t = 1.29 s it will  reach the highest point, at this point speed is zero.

V0 = V(t) - a * t

V0 = 0 - (-9.81) * 1.29 = 12.65 m/s.

Now we can go back to the other equation:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = X(t) - V0 * t - 1/2 * a * t^2

X0 = 0 - 12.65 * 6.3 - 1/2 * (-9.81) * 6.3^2 = 115 m