A baseball is seen to pass upward by a window 24 m above the street with a vertical speed of 8 m/s. The ball was thrown from the street.
(a) What was its initial speed?
(b) What altitude does it reach?
(c) How long after it was thrown did it pass the window?
(a) Try using conservation of energy
Mg*(24 m)+(1/2)M*(8 m/s)^2 = (1/2) M Vo^2
Cancel out the M's and solve for Vo.
2) Vo^2/(2 g)
3) How long does it take the velocity to go from Vo to 8 m/s?
(Vo -8)= g t
Solve for t
To answer these questions, we can use the equations of motion for projectile motion.
First, let's identify the given information:
- The vertical displacement (altitude) of the baseball is 24 m.
- The vertical velocity of the baseball is 8 m/s.
(a) What was its initial speed?
To find the initial speed, we need to calculate the magnitude of the initial velocity. The vertical velocity is equal to the initial velocity since the baseball is thrown vertically.
Given:
Vertical velocity (Vy) = 8 m/s
The magnitude of the initial velocity (V₀) can be calculated using the formula:
V₀ = √(Vx² + Vy²)
Assuming there is no horizontal velocity (Vx = 0), the equation simplifies to:
V₀ = √(0 + 8²) = √64 = 8 m/s
Therefore, the initial speed of the baseball is 8 m/s.
(b) What altitude does it reach?
To find the maximum altitude reached by the baseball, we need to find the time it takes to reach that point. At the maximum height, the vertical velocity becomes zero.
Using the formula:
Vf = Vi + at
Given:
Final velocity (Vf) = 0 m/s (at maximum height)
Initial velocity (Vi) = 8 m/s
Acceleration (a) = -9.8 m/s² (acceleration due to gravity, directed downwards)
0 = 8 - 9.8t
Solving for time (t):
9.8t = 8
t = 8 / 9.8
t ≈ 0.82 s
Now, we can find the maximum altitude using the formula:
Δy = Vit + 0.5at²
Given:
Initial velocity (Vi) = 8 m/s
Time (t) = 0.82 s
Acceleration (a) = -9.8 m/s²
Δy = (8)(0.82) + 0.5(-9.8)(0.82)²
Δy = 6.56 + 0.5(-9.8)(0.6724)
Δy ≈ 6.56 - 3.268
Δy ≈ 3.292 m
Therefore, the baseball reaches a maximum altitude of approximately 3.292 m.
(c) How long after it was thrown did it pass the window?
To find the time when the baseball passes the window, we can use the formula:
Δy = Vit + 0.5at²
Given:
Initial velocity (Vi) = 8 m/s
Vertical displacement (Δy) = 24 m
Acceleration (a) = -9.8 m/s²
Rearranging the equation:
0.5at² + Vit - Δy = 0
Solving this quadratic equation for time (t), we get two solutions. We'll take the positive one since time cannot be negative.
t = (-Vi + √(Vi² - 4(0.5a)(-Δy))) / (2(0.5a))
t = (-8 + √(8² - 4(0.5)(-9.8)(-24))) / (2(0.5)(-9.8))
t ≈ (-8 + √(64 - 4(0.5)(-9.8)(-24))) / (-9.8)
t ≈ (-8 + √(64 - 2358.72)) / (-9.8)
t ≈ (-8 + √(2358.72 - 64)) / (-9.8)
t ≈ (-8 + √(2294.72)) / (-9.8)
t ≈ (-8 + 47.90) / (-9.8)
t ≈ 39.90 / -9.8
t ≈ -4.074
Since time cannot be negative, we disregard this solution and look for the positive one.
t ≈ 4.074
Therefore, approximately 4.074 seconds after it was thrown, the baseball passes the window.