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:

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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.

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

answerscientific theory;