A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 65 m horizontally away from and below the point of release. What is the speed of the ball just before it strikes the ground?
A. 43m/s.
B. 49m/s.
C. 39m/s.
D. 36m/s.
E. 14m/s.

Just before it strikes the ground, the speed of the ball is 46.4 m/s.

Explanation:

Hi there!

The equation of position and velocity of the ball at a time t are the following:

r  = (x0 + v0 · t, h0 + 1/2 · g · t²)

v = (v0, g · t)

Where:

r = position vector of the ball at a time t.

x0 = initial horizontal position of the ball.

v0 = initial velocity.

t = time.

h0 = initial height.

g = acceleration due to gravity.

v = velocity of the ball at a time t.

Let´s place the origin of the frame of reference on the ground at the edge of the building. The initial position of the ball is r0 = (0, 100) m.

Then, the final position of the ball will be r1 = (65, 0) m.

Using the equation of the vertical component of the position vector, we can find the time at which the ball hits the ground (height = 0  m):

h =  h0 + 1/2 · g · t²

0 m = 100 m - 1/2 · 9.8 m/s² · t²

-100 m / -4.9 m/s² = t²

t = 4.5 s

We know that in 4.5 s the ball travels 65 m in the horizontal direction. Then, using the equation of the horizontal component of the position vectro, we can find the initial horizontal velocity, v0:

x = x0 + v0 · t   (x0 = 0)

x = v0 · t

x/t = v0

65 m / 4.5 s = v0

v0 = 14.4 m/s

Now, let´s calculate the vertical component of the velocity vector at t = 4.5 s:

vy = g · t

vy = -9.8 m/s² · 4.5s

vy = -44.1 m/s

When the ball hits the ground, its velocity vector will be:

v = (14.4, -44.1) m/s

The magnitude of this vector is calculated as follows:

|v| = √[(14.4 m/s)² + (-44.1 m/s)²] = 46.4 m/s

Just before it strikes the ground, the speed of the ball is 46.4 m/s.

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