A 3rd baseman is throwing a ball to 1st base which is 39.0m away. It leaves his hand at 38.0m/s at a height of 1.50m from the ground and makes an angle of 20 Degrees. How high will it be when it gets to 1st base.
Vo = 38 m/s @ 20o.
Xo = 38*cos20 = 35.71 m/s.
Yo = 38*sin20 = 13.0 m/s.
Xo * Tr = 39 m.
35.71*Tr = 39
Tr = 1.09 s. = Time to reach 39 meters.
h = ho + Yo*Tr + 0.5g*Tr^2.
h = 1.5 + (13*1.09 - 4.9(1.090)^2)=9.85
Meters above gnd when it reaches 1st base.
. Which of the following sentences is a Normative Claim? Discuss why.
a. The third baseman dropped the ball.
b. The outfielder hit a homerun.
c. The second baseman should not have dropped the ball.
d. The pitcher committed an error.
To determine how high the ball will be when it reaches first base, we need to find the vertical component of its displacement. We can use the following equation:
y = y0 + (v0 * sinθ) * t - (0.5 * g * t^2),
where:
- y is the vertical displacement,
- y0 is the initial height (1.50m),
- v0 is the initial velocity (38.0m/s),
- θ is the launch angle (20 degrees),
- t is the time of flight, and
- g is the acceleration due to gravity (-9.8 m/s^2).
First, let's find the time of flight by using the equation:
y = y0 + (v0 * sinθ) * t - (0.5 * g * t^2).
Since the ball starts and ends at the same height (1.50m), we can rewrite the equation as:
0 = (v0 * sinθ) * t - (0.5 * g * t^2).
Rearranging the equation, we get:
0.5 * g * t^2 = (v0 * sinθ) * t.
Now, we can solve for t:
0.5 * g * t = v0 * sinθ.
t = (2 * v0 * sinθ) / g.
Substituting the given values into the equation:
t = (2 * 38.0 * sin(20)) / (-9.8).
t ≈ 3.03 seconds.
Now that we have the time of flight, we can calculate the vertical displacement by using the equation:
y = y0 + (v0 * sinθ) * t - (0.5 * g * t^2).
Let's substitute the values:
y = 1.50 + (38.0 * sin(20)) * 3.03 - (0.5 * (-9.8) * (3.03)^2.
y ≈ 11.8 meters.
Therefore, the ball will be approximately 11.8 meters high when it reaches first base.
To find out how high the ball will be when it reaches 1st base, we need to use the principles of projectile motion.
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The vertical component of the initial velocity (v₀y) can be calculated using the formula:
v₀y = v₀ * sin(θ)
where v₀ is the magnitude of the initial velocity (38.0 m/s) and θ is the angle of the throw (20 degrees).
v₀y = 38.0 * sin(20)
Step 2: Calculate the time it takes for the ball to reach first base.
The time (t) can be found using the equation:
t = distance / horizontal velocity
Since the horizontal velocity doesn't change during the flight, it can be found by multiplying the initial velocity (v₀) by the cosine of the angle:
v₀x = v₀ * cos(θ)
t = 39.0 / (38.0 * cos(20))
Step 3: Calculate the vertical displacement.
The vertical displacement (Δy) can be found using the equation:
Δy = v₀y * t - 0.5 * g * t²
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Δy = (38.0 * sin(20)) * t - 0.5 * 9.8 * t²
Step 4: Add the initial height to the vertical displacement.
To find the height when the ball reaches 1st base, we need to add the initial height (1.50 m) to the vertical displacement.
Height = Initial height + Δy
Height = 1.50 + Δy
By substituting the values calculated in the previous steps, you can find the final height when the ball reaches 1st base.