Question

A long jumper reaches a horizontal velocity of 6ms just prior to leaving the ground. If she jumps a distance of 6 m from her starting point, modeling what was the highest point in her trajectory? Can you please show all work? I have tried this problem repeatedly and cannot seem to get to the answer given as 1.23 meters

Answer 1

Range , which is horizontal distance = v² sin(2$\Theta$)/g = 6 m

2vsin$\Theta$ vcos$\Theta$/g = 6

v cos$\Theta$ = 6 m/s (given, horizontal velocity)

2v sin$\Theta$ * 6 /9.8 = 6

v sin$\Theta$ = 4.9 m/s

Maximum height = (vsin$\Theta$)²/2g = 4.9*4.9/19.6 = 1.225 m which is approximately 1.23 meters