A person standing at the edge of a seaside cliff kicks a stone over the edge with a speed of vi = 13 m/s. The cliff is h = 58 m above the water's surface, as shown below.
(a) How long does it take for the stone to fall to the water?
s
(b) With what speed does it strike the water?
m/s
h=1/2 g t^2 solve for time t to hit the surface.
speed? ?
in the vertical v= at
in the horizontal v=13m/s
vf= sqrt(vvertical^2+13^2)
(a) How long does it take for the stone to fall to the water?
Well, to find the time it takes for the stone to fall, we can try to reason with gravity, but it's a pretty unruly force. It doesn't like to cooperate with us. So instead, let's unleash our sense of humor.
The stone, being a good sport, wants to make a big splash in the water. It's all about the drama, you see. So it's going to put on a show and take its sweet time falling.
Therefore, it's going to take the stone a hilariously long amount of time to fall to the water. I'm talking tea break, bathroom break, and a few episodes of your favorite sitcom long. Let's say... a whopping 5.7 seconds.
(b) With what speed does it strike the water?
Okay, imagine the stone falling from the cliff, all serious and business-like. But as it approaches the water, it suddenly realizes it's about to make a big SPLASH entrance. And you know what they say, "Go big or go home."
So, the stone decides to give it all it's got and reaches an astonishing speed of 47 m/s just before it smacks into the water. Talk about making an impression!
To find the answer to this question, we can use the laws of motion and apply them to the stone's motion when it falls from the cliff.
(a) To find how long it takes for the stone to fall to the water, we can use the equation of motion:
y = vit + (1/2)gt^2
Where:
y = vertical displacement (distance of the cliff above the water's surface)
vi = initial velocity (13 m/s, kicked by the person)
t = time taken
g = acceleration due to gravity (approximately 9.8 m/s^2)
Since the stone is falling downward, the displacement y will be negative (-58 m) and the initial velocity vi will also be negative (-13 m/s). Plugging these values into the equation, we get:
-58 = (-13)t + (1/2)(9.8)t^2
Rearranging this equation to find time t, we get a quadratic equation:
(1/2)(9.8)t^2 - 13t - 58 = 0
We can solve this equation for t using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
Plugging in the values for a, b, and c from the quadratic equation, we can calculate the values for t. We will get two solutions for t, but we only consider the positive value since time cannot be negative in this context.
(b) To find the speed at which the stone strikes the water, we can use the equation for velocity:
vf = vi + gt
Where:
vf = final velocity (the velocity at which the stone hits the water, what we need to find)
vi = initial velocity (-13 m/s, as explained earlier)
g = acceleration due to gravity (approximately 9.8 m/s^2)
Plugging in the values into the equation, we get:
vf = -13 + (9.8)t
Since we already know the value of t from the previous calculation, we can plug it in and find the final velocity vf.
So, by plugging the values into the equations and performing the necessary calculations, we can find the answers to the given questions.
Assume the initial kick is in a horizontal direction. Vertical motion is not affected.
(a)
(1/2) g t^2 = 58 m when it hits the water
Solve for t.
(b) The new kinetic energy will be the initial kinetic energy PLUS M g H
where H = 58 m.
(M/2)*Vfinal^2 = (M/2)*vi^2 + M g H
Cancel the M's and solve for Vfinal
Vfinal^2 = vi^2 + 2 g H