Projectile Motion – Launching a Ball from Height
Question:
A ball is launched from a height of 5 m above the ground at an initial speed of 10 m/s. Which ONE of the following statements BEST describes the conditions for the ball to reach the ground at the highest speed (neglecting air friction)?
(A) Launching the ball directly upwards.
(B) Launching the ball at 45° upwards from the horizontal.
(C) Launching the ball horizontally.
(D) Launching the ball directly downwards.
(E) Any of the above.
Answer with Detailed Explanation:
To determine the final speed of the ball when it hits the ground, we apply the principle of conservation of mechanical energy. In the absence of air resistance, the total mechanical energy (kinetic + potential) of the ball remains constant throughout its motion.
At the point of launch:
v = 10 m/s (initial speed),
h = 5 m (initial height),
g = 9.8 m/s² (acceleration due to gravity).
At the moment the ball hits the ground, its potential energy is zero, and all the initial mechanical energy has converted to kinetic energy:
Setting Ei = Ef gives:
We can cancel m from both sides:
Solving:
(1/2)(100) + 49 = (1/2)vf²
50 + 49 = (1/2)vf²
99 = (1/2)vf² ⇒ vf² = 198 ⇒ vf ≈ 14.07 m/s
The final speed does not depend on the launch direction. Whether the ball is thrown directly up, horizontally, at an angle, or straight down, as long as its initial speed and height are the same, the final speed when hitting the ground will also be the same.