10. A baseball is hit with a speed of 27.0m/s at an angle of 45.0^circ It lands on the flat roof of a 13.0-m-tall nearby build-ing. If the ball was hit when it was 1 .0 m above the ground , what horizontal distance does it travel before it lands on the building?

To solve this problem, we need to find the time it takes for the baseball to reach the height of the building and then use that time to calculate the horizontal distance traveled.<br /><br />Given:<br />- Initial speed of the baseball, \( v = 27.0 \, \text{m/s} \)<br />- Angle of projection, \( \theta = 45.0^\circ \)<br />- Height of the building, \( h_{\text{building}} = 13.0 \, \text{m} \)<br />- Initial height of the ball, \( h_{\text{initial}} = 1.0 \, \text{m} \)<br />- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)<br /><br />First, we calculate the vertical component of the initial velocity:<br />\[ v_y = v \sin(\theta) = 27.0 \sin(45.0^\circ) = 27.0 \times \frac{\sqrt{2}}{2} = 27.0 \times 0.707 = 19.0 \, \text{m/s} \]<br /><br />Next, we find the time it takes for the ball to reach the height of the building:<br />\[ h_{\text{building}} - h_{\text{initial}} = v_y t + \frac{1}{2} g t^2 \]<br />\[ 13.0 - 1.0 = 19.0 t + \frac{1}{2} \times 9.8 \times t^2 \]<br />\[ 12.0 = 19.0 t + 4.9 t^2 \]<br /><br />This is a quadratic equation:<br />\[ 4.9 t^2 + 19.0 t - 12.0 = 0 \]<br /><br />We solve this using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):<br />\[ a = 4.9, \, b = 19.0, \, c = -12.0 \]<br />\[ t = \frac{-19.0 \pm \sqrt{19.0^2 - 4 \times 4.9 \times (-12.0)}}{2 \times 4.9} \]<br />\[ t = \frac{-19.0 \pm \sqrt{361 + 235.2}}{9.8} \]<br />\[ t = \frac{-19.0 \pm \sqrt{596.2}}{9.8} \]<br />\[ t = \frac{-19.0 \pm 24.4}{9.8} \]<br /><br />We take the positive root:<br />\[ t = \frac{5.4}{9.8} \approx 0.55 \, \text{s} \]<br /><br />Now, we calculate the horizontal distance traveled:<br />\[ x = v \cos(\theta) t = 27.0 \cos(45.0^\circ) t = 27.0 \times \frac{\sqrt{2}}{2} \times 0.55 \]<br />\[ x = 27.0 \times 0.707 \times 0.55 \approx 10.0 \, \text{m} \]<br /><br />Therefore, the horizontal distance the baseball travels before it lands on the building is approximately \( 10.0 \, \text{m} \).