A girl jumps off a diving board that is 3m high. If she jumps of the board at 8m/s How high does she get in the air? square At what velocity does she hit the water? square

To solve this problem, we need to use the kinematic equations of motion under constant acceleration due to gravity.<br /><br />Given information:<br />- The diving board is 3 m high.<br />- The girl jumps off the board at a velocity of 8 m/s.<br /><br />Step 1: Calculate the maximum height the girl reaches in the air.<br />We can use the kinematic equation:<br />h = (v^2) / (2g)<br />Where:<br />h = maximum height<br />v = initial velocity<br />g = acceleration due to gravity (approximately 9.8 m/s^2)<br /><br />Substituting the given values:<br />h = (8 m/s)^2 / (2 × 9.8 m/s^2)<br />h = 64 m^2/s^2 / 19.6 m/s^2<br />h = 3.27 m<br /><br />Therefore, the girl reaches a maximum height of 3.27 m in the air.<br /><br />Step 2: Calculate the velocity at which the girl hits the water.<br />We can use the kinematic equation:<br />v_f^2 = v_i^2 + 2as<br />Where:<br />v_f = final velocity<br />v_i = initial velocity<br />a = acceleration (due to gravity)<br />s = displacement (downward distance)<br /><br />Since the girl starts from the diving board and falls 3 m, the displacement is -3 m (downward).<br />Substituting the given values:<br />v_f^2 = (8 m/s)^2 + 2 × (-9.8 m/s^2) × (-3 m)<br />v_f^2 = 64 m^2/s^2 + 58.8 m^2/s^2<br />v_f^2 = 122.8 m^2/s^2<br />v_f = √(122.8 m^2/s^2)<br />v_f = 11.08 m/s<br /><br />Therefore, the girl hits the water at a velocity of 11.08 m/s.