A student drops a vail from a height of 70.0m.If the ball remains in the air for 3.78s, determine all unknowns and answer the following question. Assume the ball's spead changes steadily. v_(0)=square unit v=square unit v_(f)=square unit v t=square unit What was the ball's speed just before striking the ground? square 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 ball is dropped from a height of 70.0 m.<br />- The ball remains in the air for 3.78 s.<br /><br />Unknowns:<br />- Initial velocity (v0)<br />- Final velocity (vf)<br />- Time (t)<br /><br />Step 1: Calculate the time (t) the ball is in the air.<br />Since the ball is dropped from rest, the initial velocity (v0) is 0 m/s.<br />Using the kinematic equation: v = v0 + at<br />Where:<br />- v is the final velocity (which is the velocity just before striking the ground)<br />- a is the acceleration due to gravity (g = 9.8 m/s²)<br />- t is the time<br /><br />Rearranging the equation, we get:<br />t = (v - v0) / a<br />t = (vf - 0) / 9.8<br />t = 3.78 s<br /><br />Step 2: Calculate the final velocity (vf) just before the ball strikes the ground.<br />Using the kinematic equation: v² = v0² + 2as<br />Where:<br />- v is the final velocity (which is the velocity just before striking the ground)<br />- v0 is the initial velocity (0 m/s)<br />- a is the acceleration due to gravity (g = 9.8 m/s²)<br />- s is the distance traveled (70.0 m)<br /><br />Rearranging the equation, we get:<br />vf = √(v0² + 2as)<br />vf = √(0² + 2 × 9.8 × 70.0)<br />vf = √(1372)<br />vf = 36.9 m/s<br /><br />Therefore, the ball's speed just before striking the ground is 36.9 m/s.