Answer to Question #252385 in Civil and Environmental Engineering for Joma

Question

A golf ball is driven horizontally from an elevated tee with a velocity of 27 m/s. It strikes the fairway 2.5 sec later. How far has it fallen horizontally?

Answer 1

$y = ut+ \frac12at²\\ y = 0(t) + \frac12(9.8)+(2.5)²\\ y = 0+ 4.9(6.25)\\ y = 30.625m$

$\therefore$ The vertical distance is 30.625m