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?
y=ut+12at2y=0(t)+12(9.8)+(2.5)2y=0+4.9(6.25)y=30.625my = ut+ \frac12at²\\ y = 0(t) + \frac12(9.8)+(2.5)²\\ y = 0+ 4.9(6.25)\\ y = 30.625my=ut+21at2y=0(t)+21(9.8)+(2.5)2y=0+4.9(6.25)y=30.625m
∴\therefore∴ The vertical distance is 30.625m
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