15. When Babe Ruth (baseball player) hit a homer over the 80-m-high right field fence 98 m from home plate, roughly what was the minimum speed of the ball when it left the bat? Assume the ball was hit 1.0 m above the ground and its path initially made a 36^circ angle with the ground.

To find the minimum speed of the ball when it left the bat, we can use the principles of projectile motion. The horizontal distance traveled by the ball is given by the formula:<br /><br />\[ d = v_0 \cdot t \]<br /><br />where \( d \) is the horizontal distance, \( v_0 \) is the initial velocity (or speed), and \( t \) is the time of flight.<br /><br />The time of flight can be calculated using the vertical motion of the ball. The vertical distance traveled by the ball is given by the formula:<br /><br />\[ y = v_0 \cdot t - \frac{1}{2} g t^2 \]<br /><br />where \( y \) is the vertical distance, \( g \) is the acceleration due to gravity (approximately 9.8 m/s²), and \( t \) is the time of flight.<br /><br />Given that the ball was hit 1.0 m above the ground and its path initially made a 36° angle with the ground, we can use the following equations to find the initial velocity:<br /><br />\[ \tan(36^\circ) = \frac{1.0}{d} \]<br /><br />\[ d = \frac{1.0}{\tan(36^\circ)} \]<br /><br />\[ d \approx 1.55 \, \text{m} \]<br /><br />Now, we can use the horizontal distance and the time of flight to find the initial velocity:<br /><br />\[ v_0 = \frac{d}{t} \]<br /><br />\[ t = \sqrt{\frac{2y}{g}} \]<br /><br />\[ t = \sqrt{\frac{2 \cdot 1.0}{9.8}} \]<br /><br />\[ t \approx 0.45 \, \text{s} \]<br /><br />\[ v_0 = \frac{98}{0.45} \]<br /><br />\[ v_0 \approx 217.78 \, \text{m/s} \]<br /><br />Therefore, the minimum speed of the ball when it left the bat was approximately 217.78 m/s.