A baseball is thrown from the roof of a 24.0-m-tall building with an initial velocity of magnitude 10.0 m/s and directed at an angle of 53.1° above the horizontal.
oh well, I suppose I can pretend to know what the question is.
v = vertical speed
Vi = initial vertical speed
so
Vi = 10 sin 53.1
How long rising?
v = Vi - g t = Vi - 9.81 t
at top, v = 0
so
tt = Vi/9.81 at the top (calling tt the top time)
How high at top?
h = 24 + Vi tt - 4.9 tt^2
when at ground? (using tg)
0 = 24 + Vi tg - 4.9 tg^2
solve quadratic for tg at ground
how far?
constant u = 10 cos 53.1
d = u * tg
To solve this problem, we can break it down into two components: the vertical motion and the horizontal motion. Let's start with the vertical motion:
1. Find the initial vertical velocity (Vi) using the given initial velocity and launch angle. Since the angle is above the horizontal, the vertical component of the initial velocity is given by Vi = V * sin(theta), where V is the initial velocity magnitude and theta is the launch angle.
Vi = 10.0 m/s * sin(53.1°)
2. Calculate the time it takes for the baseball to reach its maximum height using the formula for vertical motion:
Vy = Vi - g * t
At the maximum height, Vy is zero. So, 0 = Vi - g * t. Rearranging the equation, we get:
t = Vi / g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
3. Substitute the known values into the equation to find the time:
t = (10.0 m/s * sin(53.1°)) / 9.8 m/s^2
Now, let's move on to the horizontal motion:
4. Find the horizontal component of the initial velocity (Vx) using the given initial velocity and launch angle. The horizontal component of the initial velocity is given by Vx = V * cos(theta), where V is the initial velocity magnitude and theta is the launch angle.
Vx = 10.0 m/s * cos(53.1°)
5. Calculate the time taken for the baseball to hit the ground. Since the horizontal velocity remains constant throughout the motion, and at the end, the displacement in the horizontal direction is equal to the initial horizontal velocity multiplied by the time taken.
We need to find the time when the baseball hits the ground, which is the time taken for it to reach a height of 0 meters again. This is twice the time calculated in step 3.
t_total = 2 * t
6. Finally, use the time taken for the ball to hit the ground to find the horizontal displacement (range) using the equation:
x = Vx * t_total
Substitute the known values into the equation to find the horizontal displacement (range) of the baseball.
To find the horizontal distance covered by the baseball when it reaches the ground, we need to break down the initial velocity into its horizontal and vertical components.
The vertical component of the initial velocity can be found by multiplying the magnitude of the velocity (10.0 m/s) by the sine of the angle (53.1°).
Vertical component = 10.0 m/s * sin(53.1°)
Similarly, the horizontal component of the initial velocity can be found by multiplying the magnitude of the velocity by the cosine of the angle.
Horizontal component = 10.0 m/s * cos(53.1°)
Now, let's calculate the vertical component:
Vertical component = 10.0 m/s * sin(53.1°)
Vertical component ≈ 10.0 m/s * 0.799 = 7.99 m/s
Next, let's calculate the time it takes for the baseball to reach the ground. We can use the equation for vertical motion under constant acceleration:
Δy = v₀y * t + (1/2) * a * t²
Since the initial vertical velocity is upwards (positive), and the final vertical displacement is downward (negative), we can set the initial position (Δy) to zero. Therefore, the equation becomes:
0 = (7.99 m/s) * t + (1/2) * (-9.8 m/s²) * t²
This is a quadratic equation that can be solved for time (t):
(1/2) * (-9.8 m/s²) * t² + (7.99 m/s) * t = 0
Simplifying the equation, we get:
-4.9 * t² + 7.99 * t = 0
Now, we can solve this equation using the quadratic formula:
t = (-b ± √(b² - 4*a*c)) / (2*a)
For this equation, a = -4.9, b = 7.99, and c = 0. Plugging these values into the formula, we get:
t = (-7.99 ± √(7.99² - 4 * -4.9 * 0)) / (2 * -4.9)
Simplifying further:
t = (-7.99 ± √(63.84)) / (-9.8)
Calculating the square root:
t = (-7.99 ± 7.992) / (-9.8)
Now we have two possible values for time, one positive and one negative. We discard the negative value because time cannot be negative:
t ≈ (0.002 s) / (-9.8)
Therefore, t ≈ 0.002 s
Now that we have the time it takes for the baseball to reach the ground, we can calculate the horizontal distance traveled using the horizontal component of the initial velocity and the time:
Horizontal distance = horizontal component * time
Horizontal distance = 10.0 m/s * cos(53.1°) * 0.002 s
Calculating the value:
Horizontal distance ≈ 10.0 m/s * 0.601 * 0.002 s
Therefore, the horizontal distance covered by the baseball when it reaches the ground is approximately 0.012 m, or 12 cm.