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 18 m/s at angle of 35 degrees 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/s2.How far above the top of fence will the ball
pass? Consider the diameter of the ball to be
negligible.
Answer in units of m
See my later answer to the same question, submitted by "Elle".
http://www.jiskha.com/display.cgi?id=1319098671
To find how far above the top of the fence the ball will pass, we can break down the problem into horizontal and vertical components.
1. Start by finding the initial velocity components:
- The horizontal component of the initial velocity (Vx) is given by V * cos(θ), where V is the initial velocity of 18 m/s and θ is the angle of 35 degrees.
- The vertical component of the initial velocity (Vy) is given by V * sin(θ).
Substitute the values:
- Vx = 18 m/s * cos(35°) = 14.753 m/s
- Vy = 18 m/s * sin(35°) = 10.245 m/s
2. Determine the time it takes for the ball to reach the fence:
- Since the horizontal distance to the fence is 20 m and the horizontal component of the velocity is 14.753 m/s, we can use the equation:
time = distance / velocity
Substitute the values:
- time = 20 m / 14.753 m/s ≈ 1.356 s
3. Calculate the vertical displacement of the ball during this time:
- The equation to calculate displacement in the vertical direction is:
y = Vy * time + (1/2) * acceleration * time^2
- We know the initial velocity in the vertical direction (Vy) is 10.245 m/s, the time is 1.356 s, and the acceleration due to gravity is -9.8 m/s^2 (negative because it acts downward).
Substitute the values:
y = 10.245 m/s * 1.356 s + (1/2) * (-9.8 m/s^2) * (1.356 s)^2 ≈ 7.912 m
4. Determine how far above the top of the fence the ball will pass:
- Since the top of the fence is 4 m high, the ball will pass 7.912 m - 4 m = 3.912 m above the top of the fence.
Therefore, the ball will pass approximately 3.912 meters above the top of the fence.
To solve this problem, we need to find the vertical distance above the top of the fence where the ball will pass.
Let's break down the problem into two components: horizontal and vertical.
For the horizontal component:
- The initial velocity of the ball is 18 m/s at an angle of 35 degrees above the horizontal.
- The horizontal distance traveled by the ball is 20 m.
To find the time it takes for the ball to travel 20 m horizontally, we can use the horizontal component of the initial velocity and the distance formula:
distance = velocity * time
Rearranging the formula, we can solve for time:
time = distance / velocity
Plugging in the values:
time = 20 m / (18 m/s * cos(35 degrees))
Calculating this, we find the time it takes for the ball to travel 20 m horizontally is approximately 1.131 seconds.
Now let's move on to the vertical component:
- The vertical distance traveled by the ball is the distance above the top of the fence.
- The initial vertical velocity of the ball is given by the vertical component of the initial velocity: velocity * sin(angle).
- The acceleration due to gravity is -9.8 m/s^2 (negative because it acts downward).
Using the equation of motion, we can find the vertical distance traveled by the ball:
distance = initial velocity * time + (1/2) * acceleration * time^2
Since the ball starts from rest vertically, the initial vertical velocity is 0.
distance = (1/2) * acceleration * time^2
Substituting the values:
distance = (1/2) * (-9.8 m/s^2) * (1.131 seconds)^2
Calculating this, we find the vertical distance above the top of the fence where the ball will pass is approximately 4.078 meters.
Therefore, the ball will pass approximately 4.078 meters above the top of the fence.