Surprisingly, very few athletes can jump more than 2 feet (0.6 m) straight up. Use d = 1/2?gt?2 and solve for the time one spends moving upward in a 0.6-m vertical jump. Then double it for the “hang time”—the time one’s feet are off the ground.
ANSWER: STEP 1:- We know the equation of motion, 2 d = 1/2 at ----------------(1) Where, d is the total distance travelled, u is the initial velocity, 2 a is the acceleration and for this case a = g = 9.8 m/s , t is the time interval. STEP 2:- According to the question, the person covered a distance of 0.6 meters. The acceleration value we know already, a = 9.8 m/s . So, let’s put the values in equation (1), 0.6 m = 1/2 × 9.8 × t t = (0.6 × 2) / 9.8 = 0.122 t = 0.122 = 0.349 s. This is the time required to reach the height of 0.6 meters. The total time with the feet above the ground is, 2 × t = 2 × 0.349 = 0.699 = 0.7s (approximately) CONCLUSION: There is a little ambiguity in the question itself. The initial speed of the jumper is essential to find the time periods. Because the time period depends on the initial velocity. The more the initial velocity, more height it would reach an d the more time it stays in air. The actual equation of motion to solve this is, 2 d = ut + 1/2 at ----------------(2) Where u is the initial velocity. Apart from that, the time intervals we calculated are 0.349 s to reach the given height and 0.7 s is the total time for which the feet of the jumper will be above the ground.