b) For the given angle Q=57^circ what is the smallest initial speed that can get the boll to the top? __ __

To find the smallest initial speed that can get the object to the top, we need to use the concept of projectile motion. The formula for the maximum height reached by an object in projectile motion is given by:<br /><br />\[ h = \frac{v_0^2 \sin^2(\theta)}{2g} \]<br /><br />where:<br />- \( h \) is the maximum height,<br />- \( v_0 \) is the initial speed,<br />- \( \theta \) is the angle of projection, and<br />- \( g \) is the acceleration due to gravity.<br /><br />Given that the angle \( Q = 57^\circ \), we can substitute this value into the formula to find the initial speed required to reach the top.<br /><br />\[ h = \frac{v_0^2 \sin^2(57^\circ)}{2g} \]<br /><br />Since we are looking for the smallest initial speed, we can assume that the maximum height \( h \) is equal to 1 unit. Rearranging the equation, we have:<br /><br />\[ v_0^2 = \frac{2gh}{\sin^2(57^\circ)} \]<br /><br />Substituting the value of \( g \) as 9.8 m/s² and solving for \( v_0 \), we get:<br /><br />\[ v_0 = \sqrt{\frac{2 \times 9.8 \times 1}{\sin^2(57^\circ)}} \]<br /><br />Calculating the value of \( v_0 \), we find that the smallest initial speed required to get the object to the top is approximately 11.95 m/s.