7. The launching speed of a certain projectile is three times the speed it has at its maximum height.Calculate the elevation angle at launching.

To solve this problem, we need to use the principles of projectile motion. Let's denote:<br /><br />- \( v_0 \) as the initial launching speed.<br />- \( v_{h} \) as the speed at maximum height.<br />- \( \theta \) as the launch angle.<br /><br />According to the problem, the launching speed \( v_0 \) is three times the speed at maximum height \( v_h \):<br /><br />\[ v_0 = 3v_h \]<br /><br />At maximum height, the vertical component of the velocity is zero, and only the horizontal component remains. The horizontal component of the velocity is given by:<br /><br />\[ v_{h} = v_0 \cos(\theta) \]<br /><br />Substituting the relationship between \( v_0 \) and \( v_h \):<br /><br />\[ v_0 = 3(v_0 \cos(\theta)) \]<br /><br />Dividing both sides by \( v_0 \) (assuming \( v_0 \neq 0 \)):<br /><br />\[ 1 = 3\cos(\theta) \]<br /><br />Solving for \( \cos(\theta) \):<br /><br />\[ \cos(\theta) = \frac{1}{3} \]<br /><br />Now, find the angle \( \theta \) using the inverse cosine function:<br /><br />\[ \theta = \cos^{-1}\left(\frac{1}{3}\right) \]<br /><br />Therefore, the elevation angle at launching is approximately \( \theta = 70.53^\circ \).