A person standing at the edge of a seaside cliff kicks a stone over the edge with a speed of 18m/s. The cliff is 52m aboce the water's surface. How long does it take for the stone to fall to the water? With what speed does it strike the water? I got how long it took, but how do i find out the speed? help please:) thanks you for your time!
The stone is kicked over the edge of the cliff (i.e. horizontally), with a horizontal speed of 18 m/s which remains constant throughout the journey. The initial vertical speed is zero. Determine v using v^2 - u^2 = 2as for vertical motion (u = 0, a = 9.8 m/s^2 and s = 52m). Final speed is the resultant of v and 18 m/s.
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To find the time it takes for the stone to fall to the water, we can use the kinematic equation:
h = (1/2)gt^2
where h is the height of the cliff, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Given that the height of the cliff is 52 meters, we can rearrange the equation to solve for t:
t = sqrt(2h / g)
t = sqrt(2 * 52 / 9.8) = sqrt(104 / 9.8) = 3.20 seconds (rounded to two decimal places)
So it takes approximately 3.20 seconds for the stone to fall to the water.
To find the speed at which the stone strikes the water, we can use the kinematic equation:
v = gt
where v is the final velocity (speed), g is the acceleration due to gravity, and t is the time.
Given that g is approximately 9.8 m/s^2 and the time t is 3.20 seconds, we can substitute these values into the equation:
v = 9.8 * 3.20 = 31.36 m/s
Therefore, the stone strikes the water with a speed of approximately 31.36 m/s.