A child throws a ball with an initial speed of 8 m/s at an angle of 40 degrees above the horizontal. The ball leaves her hand 1 m above the ground. How long is the ball in flight before it hits the ground? What is the magnitude of the ball's velocity just before it hits the ground? At what angle below the horizontal does the ball approach the ground?
it falls
initial speed up = Vi = 8 sin 40
y = yo + Vi t -4.9 t^2
0 = 1 + 8 sin 40 t -4.9 t^2
solve for t, time to hit ground
horizontal velocity u = 8 cos 40 forever
find v at t = time to hit ground
v = Vi -9.8 t (will be negative, down)
tan angle = v/u
magnitude speed = sqrt(u^2+v^2)
When it says magnitude of velocity is that the same as the final velocity before impact? The word "magnitude" threw me off.
magnitude is the size sqrt (u^2+v^2)
To solve this problem, we can use the principles of projectile motion. We need to break the initial velocity into its horizontal and vertical components.
1. Finding the time of flight:
- The initial vertical velocity (v₀y) can be found using the formula: v₀y = v₀ * sin(θ)
- Here, v₀ is the initial velocity (8 m/s) and θ is the angle above the horizontal (40 degrees).
- v₀y = 8 m/s * sin(40°) ≈ 5.14 m/s
- The time of flight (t) can be calculated using the formula: t = 2 * v₀y / g
- Here, g represents the acceleration due to gravity, which is approximately 9.8 m/s².
- t = 2 * 5.14 m/s / 9.8 m/s² ≈ 1.05 seconds
- So, the ball is in flight for approximately 1.05 seconds before it hits the ground.
2. Finding the velocity just before hitting the ground:
- The final vertical velocity (vf)y just before hitting the ground will be equal to the initial vertical velocity (v₀y) because the only force acting on the ball is gravity.
- vf = v₀y = 5.14 m/s
- The horizontal component of velocity remains constant since there are no horizontal forces acting on the ball.
- So, the final horizontal velocity (vf)x will be the same as the initial horizontal velocity (v₀x).
- The initial horizontal velocity (v₀x) can be found using the formula: v₀x = v₀ * cos(θ)
- v₀x = 8 m/s * cos(40°) ≈ 6.11 m/s
- The magnitude of the ball's velocity just before hitting the ground can be found using the Pythagorean theorem:
- v = √((vf)x² + (vf)y²)
- v = √((6.11 m/s)² + (5.14 m/s)²) ≈ 8.01 m/s
- So, the magnitude of the ball's velocity just before hitting the ground is approximately 8.01 m/s.
3. Finding the angle below the horizontal:
- The angle below the horizontal at which the ball approaches the ground can be found using the inverse tangent (arctan) function:
- tan(θ) = [(vf)y / (vf)x]
- θ = arctan[(vf)y / (vf)x]
- θ = arctan[(5.14 m/s) / (6.11 m/s)] ≈ 42.6°
- So, the angle at which the ball approaches the ground is approximately 42.6 degrees below the horizontal.