4. A golf ball is hit from the ground and moves with an initial velocity of 30ms^-1 at an angle of 37^0 (a) What is the maximum height reached by the ball? (b) What is the range (horizontal distance) of the ball when it touches the ground?

(a) To find the maximum height reached by the ball, we can use the kinematic equation:<br /><br />\[ h = \frac{v_i^2 \sin^2(\theta)}{2g} \]<br /><br />where:<br />- \( h \) is the maximum height,<br />- \( v_i \) is the initial velocity,<br />- \( \theta \) is the angle of projection,<br />- \( g \) is the acceleration due to gravity.<br /><br />Given:<br />- \( v_i = 30 \, \text{m/s} \)<br />- \( \theta = 37^\circ \)<br />- \( g = 9.81 \, \text{m/s}^2 \)<br /><br />Substituting the values:<br /><br />\[ h = \frac{(30)^2 \sin^2(37^\circ)}{2 \times 9.81} \]<br /><br />\[ h = \frac{900 \times 0.384}{19.62} \]<br /><br />\[ h = \frac{345.6}{19.62} \]<br /><br />\[ h \approx 17.5 \, \text{m} \]<br /><br />So, the maximum height reached by the ball is approximately \( 17.5 \, \text{m} \).<br /><br />(b) To find the range (horizontal distance) of the ball when it touches the ground, we can use the kinematic equation:<br /><br />\[ R = \frac{v_i^2 \sin(2\theta)}{g} \]<br /><br />where:<br />- \( R \) is the range,<br />- \( v_i \) is the initial velocity,<br />- \( \theta \) is the angle of projection,<br />- \( g \) is the acceleration due to gravity.<br /><br />Given:<br />- \( v_i = 30 \, \text{m/s} \)<br />- \( \theta = 37^\circ \)<br />- \( g = 9.81 \, \text{m/s}^2 \)<br /><br />Substituting the values:<br /><br />\[ R = \frac{(30)^2 \sin(2 \times 37^\circ)}{9.81} \]<br /><br />\[ R = \frac{900 \times 0.848}{9.81} \]<br /><br />\[ R = \frac{763.2}{9.81} \]<br /><br />\[ R \approx 77.8 \, \text{m} \]<br /><br />So, the range (horizontal distance) of the ball when it touches the ground is approximately \( 77.8 \, \text{m} \).