To solve for the time when the baseball will hit the ground after being hit at an angle, we can use kinematic equations. The vertical motion of the baseball can be described by the following equation for vertical displacement:
\[ y(t) = y_0 + v_{0y} t - \frac{1}{2} g t^2 \]
Where:
- \(y(t)\) is the vertical position of the baseball at time \(t\),
- \(y_0\) is the initial height (4.8 ft),
- \(v_{0y}\) is the initial vertical velocity,
- \(g\) is the acceleration due to gravity (32 ft/s²),
- \(t\) is the time in seconds.
- Calculate the initial vertical velocity (\(v_{0y}\)):
The initial velocity can be split into vertical and horizontal components using trigonometric functions. The vertical component is given by:
\[ v_{0y} = v_0 \sin(\theta) \]
where:
- \(v_0 = 150\) ft/s (initial velocity),
- \(\theta = 52°\) (angle of projection).
Calculating \(v_{0y}\):
\[ v_{0y} = 150 \sin(52°) \]
Using a calculator to compute \(\sin(52°)\):
\[ \sin(52°) \approx 0.7880 \]
So,
\[ v_{0y} = 150 \times 0.7880 \approx 118.2 \text{ ft/s} \]
- Set up the equation for vertical displacement:
We set \(y(t)\) to 0 when the ball hits the ground:
\[ 0 = 4.8 + 118.2 t - \frac{1}{2} (32) t^2 \]
This simplifies to:
\[ 0 = 4.8 + 118.2 t - 16 t^2 \]
Rearranging gives us:
\[ 16 t^2 - 118.2 t - 4.8 = 0 \]
- Solve the quadratic equation using the quadratic formula:
The quadratic formula is:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \(a = 16\), \(b = -118.2\), and \(c = -4.8\).
Calculating the discriminant:
\[ b^2 - 4ac = (-118.2)^2 - 4 \cdot 16 \cdot (-4.8) \]
\[ = 13968.24 + 307.2 \approx 14275.44 \]
Now calculating \(t\):
\[ t = \frac{118.2 \pm \sqrt{14275.44}}{32} \]
Calculating \(\sqrt{14275.44}\):
\[ \sqrt{14275.44} \approx 119.4 \]
Now we plug in the values:
\[ t = \frac{118.2 \pm 119.4}{32} \]
Calculating the two possible values:
-
\(t = \frac{118.2 + 119.4}{32} \approx \frac{237.6}{32} \approx 7.43\)
-
\(t = \frac{118.2 - 119.4}{32} \approx \frac{-1.2}{32} \text{(not a valid solution)}\)
Thus, the only valid time when the baseball hits the ground is:
\[ t \approx 7.43 \text{ seconds} \]
So, the answer is:
7.43 seconds