8. A ball thrown horizontally at 12.2m/s from the roof of a building lands 21 .om from the base of the,building.How high is the building?

To find the height of the building, we need to use the principles of projectile motion. Since the ball is thrown horizontally, the horizontal and vertical motions are independent of each other.<br /><br />First, let's find the time it takes for the ball to land. We can use the horizontal motion to do this. The horizontal distance traveled is given by:<br /><br />\[ \text{distance} = \text{speed} \times \text{time} \]<br /><br />Rearranging for time, we get:<br /><br />\[ \text{time} = \frac{\text{distance}}{\text{speed}} \]<br /><br />Plugging in the values:<br /><br />\[ \text{time} = \frac{21 \, \text{m}}{12.2 \, \text{m/s}} \]<br /><br />\[ \text{time} \approx 1.72 \, \text{s} \]<br /><br />Now, we use the vertical motion to find the height of the building. The vertical distance (height of the building) is given by:<br /><br />\[ \text{height} = \frac{1}{2} \times \text{acceleration} \times \text{time}^2 \]<br /><br />Using the acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \):<br /><br />\[ \text{height} = \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (1.72 \, \text{s})^2 \]<br /><br />\[ \text{height} = \frac{1}{2} \times 9.8 \times 2.96 \]<br /><br />\[ \text{height} \approx 14.5 \, \text{m} \]<br /><br />Therefore, the height of the building is approximately 14.5 meters.