Question

An object is thrown vertically upward with a speed of 14.62mph. How much time from launch will it take to reach the bottom of a well 10.33 m deep?

Answer 1

Notation

u = initial speed,

v = final speed

t = time

a = acceleration due to gravity

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u = 14.62 mph = 6.5357 m/s

At maximum height , v = 0 m/s

v = u + at1

t1 = u / a

t1 = 6.5357 / 9.8

t1 = 0.6669 sec

maximum height reached, h = u² / 2a = 2.179 m

Now, considering the object is under fall, it has to fall a total distance of 2.179 + 10.33 = 12.509 m

so,

t2 = sqrt (2h / a)

t2 = sqrt (2 * 12.509 / 9.8)

t2 = 1.5977 seconds

to,

total time taken = t1 + t1

T = 2.2646 sec