This is a projectile motion question.
A ball with a mass of .4 kg, initially at rest is kicked directly toward a 3 meter fence from a point 20 meters away. The initial velocity is 17 m/s at an angle of 35 degrees. The ball is hit by nothing and there is no wind resistance, but gravity is still its normal acceleration of -9.8 m/s. How long does it take to reach the plane of the fence,how far above the top of the fence does it pass (diameter disregarded) and what is the vertical component of the velocity when it crosses the plane.
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To solve this projectile motion problem, we can break it down into horizontal and vertical components.
First, let's find the time it takes for the ball to reach the plane of the fence:
1. Separate the initial velocity into horizontal and vertical components:
- Horizontal component: Vx = V * cosθ, where V is the initial velocity and θ is the angle.
- Vertical component: Vy = V * sinθ.
In this case, V = 17 m/s and θ = 35 degrees.
So, Vx = 17 * cos(35°) and Vy = 17 * sin(35°).
2. Calculate the time of flight (T) using the vertical component:
Use the kinematic equation: y = Vy * t + (1/2) * a * t^2,
where y is the vertical distance, t is the time, and a is the acceleration (gravity).
Since the ball starts at rest in the vertical direction (Vy initial velocity = 0),
we can simplify the equation to: y = (1/2) * a * t^2.
In this case, y = -3 meters (negative since it's below the starting point) and a = -9.8 m/s^2.
So, -3 = (1/2) * -9.8 * t^2.
Solve for t by rearranging the equation and taking the square root:
t = √(2 * y / a).
Plugging in the values, t = √(2 * -3 / -9.8).
3. Find the positive value of t since time cannot be negative:
t = √(2 * 3 / 9.8).
Therefore, it takes approximately 0.78 seconds for the ball to reach the plane of the fence.
Next, let's determine how far above the top of the fence the ball passes (disregarding diameter):
1. Calculate the vertical distance covered using the time found:
Use the equation: y = Vy * t + (1/2) * a * t^2,
where y is the vertical distance, Vy is the vertical component of velocity, t is the time, and a is the acceleration.
In this case, Vy = 17 * sin(35°) and t = 0.78 seconds.
So, y = Vy * t + (1/2) * (-9.8) * t^2.
Solve for y:
y = (17 * sin(35°) * 0.78) + (1/2) * (-9.8) * (0.78)^2.
2. Subtract the height of the fence (3 meters) from the value of y obtained.
So, the ball passes approximately 0.7 meters above the top of the fence.
Finally, let's find the vertical component of the velocity when the ball crosses the plane:
1. Calculate the final vertical velocity (Vfy) using the initial vertical velocity and acceleration:
Use the equation: Vfy = Vy + a * t.
In this case, Vy = 17 * sin(35°), a = -9.8 m/s^2, and t = 0.78 seconds.
Vfy = 17 * sin(35°) + (-9.8) * 0.78.
Therefore, the vertical component of the velocity when the ball crosses the plane is approximately -5.03 m/s.