To find the acceleration of the baseball when it is thrown straight down, we first need to consider the forces acting on it. The two main forces acting on the baseball are:
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The gravitational force (weight), which can be calculated using the formula: \[ F_{\text{gravity}} = m \cdot g \] where \( m = 0.145 , \text{kg} \) (mass of the baseball) and \( g = 9.81 , \text{m/s}^2 \) (acceleration due to gravity).
So, \[ F_{\text{gravity}} = 0.145 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 1.42145 , \text{N} \approx 1.42 , \text{N} \]
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The drag force acting against the motion, which is given as \( F_{\text{drag}} = 0.4 , \text{N} \).
Now, we can determine the net force acting on the baseball using the following equation: \[ F_{\text{net}} = F_{\text{gravity}} - F_{\text{drag}} \] Substituting the values: \[ F_{\text{net}} = 1.42 , \text{N} - 0.4 , \text{N} = 1.02 , \text{N} \]
Next, we can find the acceleration \( a \) using Newton's second law \( F = m \cdot a \): \[ a = \frac{F_{\text{net}}}{m} \] Substituting the values: \[ a = \frac{1.02 , \text{N}}{0.145 , \text{kg}} \approx 7.03 , \text{m/s}^2 \]
Thus, the acceleration of the baseball when thrown straight down at that particular speed is approximately \( 7.03 , \text{m/s}^2 \).
None of the provided options (1.8 N, 65 N, 11.6 N, 0.6 N) directly describe the calculated acceleration (which should be in \( \text{m/s}^2\) not \( N \) since acceleration is not measured in Newtons). The output provided in the options likely contains an oversight since they generally should be acceleration values.
Please confirm if the options were to be in \( N \) or refer back to the context. Nonetheless, the calculated acceleration is \( \approx 7.03 , \text{m/s}^2 \).