A baseball is hit 1.3 m above home plate. The bat sends it off at an angle of 30 degrees above the horizontal at a speed of 45.0 m/s. The outfield fence is 100 m away and 11.3 m high. Does the ball clear the fence?
Vertical vi = 45 sin 30 = 22.5
horizontal u = 45 cos 30 which is constant
h = hi + vi t - 4.9 t^2
d = u t
100 = 45 cos 30 * t
so
t = (100/45)/cos 30
then
h = 1.3 + 22.5 t - 4.9 t^2
To determine if the ball clears the fence, we need to calculate the horizontal distance the ball travels and its maximum height.
1. Horizontal distance:
The horizontal distance (range) traveled by the ball can be calculated using the projectile motion formula:
Range = (velocity × time) × cos(angle)
Given:
Velocity (V) = 45.0 m/s
Angle (θ) = 30 degrees
First, convert the angle from degrees to radians:
θ_radians = θ × π / 180
θ_radians = 30 × π / 180
θ_radians = 0.5236 radians
Now, substitute the values into the formula to find the range:
Range = (45.0 m/s × time) × cos(0.5236 radians)
2. Maximum height:
The maximum height reached by the ball can be calculated using the formula for vertical motion. In this case, the initial vertical velocity is given as 0 because the initial velocity is purely horizontal.
Maximum height = (initial vertical velocity × time) - (1/2 × acceleration × time²)
Acceleration due to gravity (g) = 9.8 m/s²
Now, let's calculate the time (t) taken for the ball to reach the maximum height:
Using the equation for vertical motion:
0 = (0 m/s) - (1/2 × 9.8 m/s² × time²)
Rearranging the equation:
4.9 m/s² × time² = 0
This equation tells us that time is 0. Therefore, the ball has not yet reached the maximum height.
Since the ball has not reached its maximum height, we cannot determine if it clears the fence or not.
To determine if the baseball clears the fence, we need to calculate its horizontal and vertical distances traveled.
First, we need to break down the initial velocity of the baseball into its horizontal and vertical components.
The vertical component of the velocity can be found using the equation:
V_y = V * sin(θ)
where V is the initial velocity of 45.0 m/s, and θ is the angle of 30 degrees.
V_y = 45.0 m/s * sin(30°)
V_y ≈ 22.5 m/s
Similarly, the horizontal component of the velocity can be found using the equation:
V_x = V * cos(θ)
V_x = 45.0 m/s * cos(30°)
V_x ≈ 38.97 m/s
Next, we can calculate the time the baseball is in the air. We will use the vertical component to determine this time.
The vertical distance traveled by the ball can be calculated using the equation:
Δy = V_y * t + (1/2) * g * t^2
where Δy is the vertical distance, V_y is the vertical component of velocity, t is the time, and g is the acceleration due to gravity (-9.8 m/s^2).
Since the ball is hit above the home plate, the initial vertical displacement Δy is 1.3 m, and the final vertical displacement is 0 (since the ball is at ground level).
0 = V_y * t + (1/2) * g * t^2
Simplifying the equation, we get:
-4.9 t^2 + 22.5 t + 1.3 = 0
We can solve this quadratic equation to find the time t. Using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a),
where a = -4.9, b = 22.5, and c = 1.3.
Calculating the discriminant (b^2 - 4ac):
discriminant = (22.5)^2 - 4 * (-4.9) * 1.3
discriminant ≈ 571.02
Since the discriminant is positive, we will have two real solutions for t.
Using the quadratic formula, we have:
t = (-22.5 ± sqrt(571.02)) / (2 * -4.9)
t ≈ 4.12 seconds or t ≈ 0.065 seconds
Since the initial time of contact is not relevant to this question, we will consider the time t ≈ 4.12 seconds.
Now, we can calculate the horizontal distance using the equation:
Δx = V_x * t
Δx = 38.97 m/s * 4.12 s
Δx ≈ 160.757 m
The ball travels approximately 160.757 m horizontally before it touches the ground.
Given that the distance to the fence is 100 m, we can see that the ball will clear the fence since it travels further than the distance to the fence.