Aball is thrown horizontally from a 90 m cliff and strikes the ground 70 m from the base, what is the initial velocity?

The initial velocity is 16 m/s

Explanation:

The position of the ball at a time t is given by the following vector:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

r = position vector at time t

x0 = initial horizontal position

v0x = initial horizontal velocity

t = time

y0 = initial vertical position

v0y = initial vertical velocity

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

When the ball reaches the ground its position vector will be "r final" (see attached figure). The x-component of this vector is 70 m and the y-component is -90 m. Notice that the origin of the frame of reference is located at the throwing point (y0 and x0 = 0).  Then, we can use the equations for the x and y-component of "r" to obtain the time of flight and the initial velocity:

Using the equation for the y-component of the vector position:

y = y0 + v0y · t + 1/2 · g · t²        (y0 and v0y = 0)

-90 m = -1/2 · 9.8 m/s² · t²

-2 · -90 m/ 9.8 m/s² = t²

t = 4.3 s

Using the equation for the x-component of the vector position:

x = x0 + v0x · t

70 m = v0x · 4.3 s

v0x = 16 m/s

The initial velocity is 16 m/s

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