A ball of mass 0.4 kg, initially at rest, is kicked directly toward a fence from a point 20 m away, as shown below.
The velocity of the ball as it leaves the kicker’s foot is 16 m/s at angle of 46◦ above the horizontal. The top of the fence is 4 m high. The ball hits nothing while in flight and air resistance is negligible. The acceleration due to gravity is 9.8 m/s^2.
Determine the time it takes for the ball to reach the plane of the fence.
Answer in units of seconds.
How far above the top of fence will the ball pass? Consider the diameter of the ball to be negligible
What is the vertical component of the velocity when the ball reaches the plane of the fence? Answer in units of m/s
Please help, this is college physics!!! thank you!
To find the time it takes for the ball to reach the plane of the fence, we can use the kinematic equation for vertical motion:
y = y0 + v0y * t - (1/2) * g * t^2
Where:
y = final vertical displacement (height of the fence, 4 m)
y0 = initial vertical displacement (0 m, since the ball starts at the ground)
v0y = initial vertical component of velocity (v0 * sin(angle))
g = acceleration due to gravity (9.8 m/s^2)
t = time
Substituting the given values into the equation, we have:
4 = 0 + (16 * sin(46)) * t - (1/2) * (9.8) * t^2
Simplifying the equation, we get a quadratic equation in terms of t:
(1/2) * (9.8) * t^2 - (16 * sin(46)) * t + 4 = 0
Now, we can solve this quadratic equation for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Where a = (1/2) * (9.8), b = -(16 * sin(46)), and c = 4.
Solving this equation will give us two possible solutions for t, but we only need the positive one since time cannot be negative.
Once we have the value for t, we can substitute it back into the equation for the vertical component of velocity:
v = v0y - g * t
Where v0y = (16 * sin(46)) and g = 9.8 m/s^2.
Again, solving for v will give us the vertical component of velocity when the ball reaches the plane of the fence.
Let's calculate these values:
Using the quadratic formula:
a = (1/2) * (9.8) = 4.9
b = -(16 * sin(46)) = -11.18
c = 4
t = (-(-11.18) ± √((-11.18)^2 - 4 * 4.9 * 4)) / (2 * 4.9)
t = (11.18 ± √(124.9424 - 78.4)) / 9.8
t = (11.18 ± √46.5424) / 9.8
Let's calculate the value inside the square root:
√46.5424 ≈ 6.82
Now we can calculate the values of t:
t1 = (11.18 + 6.82) / 9.8 ≈ 1.83 seconds
t2 = (11.18 - 6.82) / 9.8 ≈ 0.44 seconds
Since time cannot be negative, we discard t2 = 0.44 seconds.
Now, substituting t = 1.83 seconds into the equation for the vertical component of velocity:
v = (16 * sin(46)) - (9.8 * 1.83)
v ≈ 9.82 m/s
Therefore, the time it takes for the ball to reach the plane of the fence is approximately 1.83 seconds, and the vertical component of velocity when the ball reaches the plane of the fence is approximately 9.82 m/s.
To summarize:
1. The time it takes for the ball to reach the plane of the fence is approximately 1.83 seconds.
2. The vertical component of velocity when the ball reaches the plane of the fence is approximately 9.82 m/s.
The horizontal velocity component, which does not change, is
Vx = Vo cos46 = 11.11 m/s. Divide the distance to the fence by Vx to obtain the flight time to the plane of the fence.
Use that time (t) and the equation for height above the ground (given below), to find out how far whether, or by how much, the ball clears the fence.
y = (16 sin46)t - (g/2)*t^2
For the vertical component of V,
Vy = 10 sin46 - gt