A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00 m above the ground. What is the magnitude of the ball's velocity just before it hits the ground?
35 m/s
To find the magnitude of the ball's velocity just before it hits the ground, we need to break down the initial velocity into its horizontal and vertical components.
Given:
Initial speed (v₀) = 8.00 m/s
Launch angle (θ) = 40.0°
Height of the ball when launched (h) = 1.00 m
First, let's find the horizontal and vertical components of the initial velocity.
The horizontal component (v₀x) remains constant throughout the ball's motion and can be found using trigonometry:
v₀x = v₀ * cos(θ)
Substituting the given values:
v₀x = 8.00 m/s * cos(40.0°)
v₀x ≈ 8.00 m/s * 0.766
v₀x ≈ 6.132 m/s
Next, let's find the vertical component of the initial velocity (v₀y). The ball's initial vertical velocity can be calculated using trigonometry:
v₀y = v₀ * sin(θ)
Substituting the given values:
v₀y = 8.00 m/s * sin(40.0°)
v₀y ≈ 8.00 m/s * 0.643
v₀y ≈ 5.144 m/s
Now, we need to calculate the time it takes for the ball to hit the ground.
The equation that relates the vertical distance traveled by the ball (h), the initial vertical velocity (v₀y), and time (t) is:
h = v₀y * t + (1/2) * g * t²
Where g is the acceleration due to gravity, approximately 9.8 m/s².
Let's solve this equation for time (t).
Rearranging the equation, we get:
(1/2) * g * t² + v₀y * t - h = 0
Now we can use the quadratic formula to solve for t:
t = (-v₀y ± sqrt(v₀y² + 2 * g * h)) / g
Plugging in the values:
t = (-(5.144 m/s) ± sqrt((5.144 m/s)² + 2 * (9.8 m/s²) * (1.00 m))) / (9.8 m/s²)
Simplifying this equation gives two possible solutions for t. Since the ball is initially launched upward, we only consider the positive root because we are interested in the time taken to reach the maximum height and then fall back to the ground.