how long did the stone fall in the air and how long did it fall in the water?

Question

stone fell from the top of a cliff into the ocean. In the air, it had an average speed of 5m/s before hitting the sea. The total distance from the top of the cliff to the sea is 200 feet and the entire fall took 15 seconds.

Answer 1

If the stone falls from an initial height of 200 with initial velocity of zero, then the formula for its height is h(t) = -(1/2)gt² where g= acceleration due to gravity = 9.8 m/s² = 32 ft/s²

It hits ground at h=0 200 -(1/2)32t² =0, solve for t: t²=200/16=2(100/16) t=(10/4)(1.414) =2.5(1.414)=3.536 seconds

If it took 15 seconds to fall to the bottom of the ocean, then it fell for 3.536 seconds in the air from cliff top to ocean top, and then for 11.464 seconds from ocean top to ocean bottom, ignoring buoyancy pressure.

Or we could have used the formula in meters per second per second, but we have to first convert 200 feet to meters. 2.54 cm in an inch. 2.54 cm times 12 inches in a foot. or 30.48 cm in a foot 100 cm=one meter

one foot = .3048 meters 200 feet = 200 times .3048 = 60.96 meters

60.96-(1/2)9.8t²=0 t²=60.96/4.9= 12.441 t= 3.527 seconds Due to rounding errors, this is not the same as 3.536 seconds. 3.54 seconds is correct to 2 decimal places, whichever method used.

That's time from cliff top to ocean top. Time from ocean top to ocean bottom is 15-3.54=11.46 seconds

Unless, the question really gave 15 seconds as the time from cliff top to ocean top.

formula for its position is h(t) = -16t²-vt+h_o where t = time, v=initial velocity and h_o= initial height

initial height = 200 feet, t= 15 second average speed = 5m/second

-16(15)² - v(15) + 200 = 0

solve for v

-16(225) -v(15) +200 = 0

v= (200-3000)/15 = - 2800/15 ft/second = 560/3 = 186 2/3 ft/sec.

If initial velocity had been zero, then the stone would have hit bottom when -16t²=-200 or t²=200/16=50/4=25/2=12.5 seconds, take the square root of 12.5 which is t=3.54 seconds. For it to take 15 seconds the initial velocity must have been upward, taking much longer for gravity to overcome. It's initial velocity was negative 560/3 feet per second or 186.667 ft./sec

It's velocity remains negative until it reaches its maximum height far above 200 feet.

take the first derivative of the equation to get -32t+560/3=0 At time t=560/3(32) = 280/48= 140/24=70/12=35/6 or 5.83 seconds. Then it falls for 5.83 seconds until reaching its initial position, then falls for another about 3.54 seconds, for a grand total time of 15 seconds. But it's rising for the almost 6 seconds, then falling for the final 9 seconds.

The average speed in the first 12 seconds is zero, as it reverses direction. Average speed in the last 3 seconds in the 200 feet fall is distance divided by time or 200/3 ft/sec.

For the first 12 seconds average speed is zero For last 3 seconds average speed is 200/3 or about 67 ft./sec. It was rising for 6 seconds, then falling for about 9 seconds.

Take 12 times zero plus 3 times 67 ft./sec = 200 and divide by 15 = average overall speed of 200/15 = 40/3 or about 13 ft/sec. The given average speed was 5 m/sec. A foot is 1/3 of a yard or less than 1/3 of a meter. Less than 1/3 of 13 is approximately the given 5 m/sec

But you're also asking how long did it fall in the water. Depends on how deep the water was and how dense the water was compared to the stone. Water slows the stone's fall, but it will be in the water until it hits bottom or some obstruction. The speed as it hits the water will be about -16(3x3)= - 144 ft/sec or less than one third that for meters per second. IF the water is 144 feet deep, it may be in the water only for a second or less.