Question

Three balls are thrown off a tall building with the same speed but in different directions. Ball A is thrown in the horizontal direction; ball B starts out at 45° above the horizontal; ball C begins its flight at 45° below the horizontal. Which ball has the greatest speed just before it hits the ground? Assume that the ground near the building is level, and ignore any effects due to air resistance.
1. Ball A

2. Ball B

3. Ball C

4. Balls B and C have the same speed, which is greater than Ball A

5. All balls have the same speed

Answer 1

Answer:

5. All balls have the same speed

Explanation:

We can solve this problem by applying considerations about the law of conservation of energy.

In fact, in absence of air resistance, the total mechanical energy of the ball (given by the sum of kinetic + potential energy) must be conserved.

When the ball is thrown from the top of the building, the mechanical energy is:

$E=KE_i + PE_i = \frac{1}{2}mu^2 + mgh$

where

m is the mass of the ball

u is the initial speed

g is the acceleration due to gravity

h is the height above the ground

Just before reaching the ground, all the mechanical energy has been converted into kinetic energy, so the new mechanical energy is:

$E=KE_f=\frac{1}{2}mv^2$

where

v is the final speed of the ball

Since the total energy is conserved, we can equate the two expression, and we can find an expression for the final speed of the ball:

$\frac{1}{2}mu^2+mgh=\frac{1}{2}mv^2\\v=\sqrt{u^2+2gh}$

We notice that the final speed of the ball depends only on:

- $u$, the initial speed of the ball, which is the same of all the three balls

- g, the acceleration due to gravity

- h, the initial height of the ball (which is the same for the three balls)

Therefore, since the final speed does not depend on the angle of projection, we can conclude that all balls have the same final speed.