A girl riding a bicycle at 2.16 m/s throws a tennis ball horizontally forward at a speed of 1.38 m/s relative to herself. The tennis ball lands 1.84 m horizontally from the point at which the girl released the ball. How high above the ground was the ball released?
To determine the height at which the ball was released, we can use the equations of motion and the concept of projectile motion.
Step 1: Identify the known quantities:
- Initial velocity of the girl on the bicycle (v₀) = 2.16 m/s (horizontal)
- Velocity of the tennis ball relative to the girl (v_b) = 1.38 m/s (horizontal)
- Horizontal distance traveled by the tennis ball (d_h) = 1.84 m
Step 2: Determine the time of flight:
Since the horizontal velocity of the ball remains constant throughout its motion, the time of flight can be calculated using the horizontal distance and the horizontal velocity relative to the girl:
Time (t) = distance (d) / velocity (v)
t = d_h / v_b
t = 1.84 m / 1.38 m/s
t ≈ 1.33 s
Step 3: Calculate the height using vertical motion equations:
In the absence of external forces, the vertical motion of the ball is subject to free-fall acceleration due to gravity (9.8 m/s²). The vertical position equation for the ball can be utilized:
Vertical distance (d_v) = (v₀ × t) + (1/2) × (g × t²)
Since the initial vertical velocity is zero (the ball is thrown horizontally), the equation simplifies to:
Vertical distance (d_v) = (1/2) × (g × t²)
d_v = 1/2 × 9.8 m/s² × (1.33 s)²
d_v ≈ 8.2 m
Therefore, the ball was released at a height of approximately 8.2 meters above the ground.