# A ball is thrown horizontally from the top of a building 49.5 m high. The ball strikes the ground at a point 43.5 m from the base of the building. Find the initial velocity of the ball. The acceleartion of gravity is 9.81 m/s/s.

First, we will calculate the time it takes for the ball to hit the ground:

To do this, we will use the equation `y = y_0 + v_0t + 1/2at^2` .

In this case, the initial velocity in the y direction is zero, as is the initial y position, and...

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First, we will calculate the time it takes for the ball to hit the ground:

To do this, we will use the equation `y = y_0 + v_0t + 1/2at^2` .

In this case, the initial velocity in the y direction is zero, as is the initial y position, and acceleration here is `9.81` `m/s^2` :

`49.5s = 1/2 * 9.81 m/s^2 * t^2`

Therefore ` t = sqrt{(2*49.5m) / (9.81m/s^2)} = 3.176s`

Now that we know t, we once again use the same equation, but this time `a=0, x_0=0` and `v_0` is the unknown: `x = v_0*t`

Therefore `v = x/t = (43.5m) / (3.176s) = 13.96 m/s`

Approved by eNotes Editorial Team