A person standing at the edge of a seaside cliff kicks a stone over the edge with a speed of vi = 20 m/s. The cliff is h = 53 m above the water's surface, as shown below.

(a) How long does it take for the stone to fall to the water?
(b) With what speed does it strike the water?

Since there is no initial vertical velocity component, the time to fall (T) is given by

H = (1/2)gT^2
H is the cliff height.
T = sqrt(2H/g)

The speed Vf at impact can be easily computed with conservation of energy.

(1/2)m Vf^2 =(1/2)m Vo^2 = mgH
The mass cancels out. Vo is the initial velocity (when kicked)

Vf^2 = Vo^2 + 2 g H

Oh i get it. You have to use the equation for an accelerating straight-line object. Thanks

(a) How long does it take for the stone to fall to the water?

Well, let me calculate this for you. But while I do, let's hope the stone doesn't look down and get scared. We wouldn't want it to wet its little pebble pants!

Using the equation s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, t is the time, and a is the acceleration, we can rearrange it to solve for t:

h = ut + (1/2)at^2

Since the stone is only affected by gravity, we have a = -9.8 m/s^2 (with a negative sign since it's acting downward). The initial velocity, u, is 20 m/s, and the distance, s, is 53 m:

53 = (20)t + (1/2)(-9.8)t^2

Now we just need to solve this quadratic equation to find t. Hold on, let me get my dusty old calculator...

(takes out a calculator, blows off some dust, and starts calculating)

Hmm... It seems like the clown academy didn't prepare me for this kind of calculation. Let me call in the math expert, Professor Calculus, for help. Professor Calculus, are you there?

(b) With what speed does it strike the water?

While we wait for Professor Calculus to arrive, let me tell you a punny water-related joke to lighten the mood: Why did the fisherman bring a ruler to the water? Because he wanted to see how long the fish he caught was!

Oh, here comes Professor Calculus! Take it away, math whiz!

Professor Calculus: Ah, thank you, Clown Bot! Let me handle the math here. To find the time it takes for the stone to fall, we can use the quadratic formula:

t = (-u ± √(u^2 - 4as)) / (2a)

Plugging in the values:

t = (-20 ± √(20^2 - 4(-9.8)(-53))) / (2(-9.8))

After some careful calculations, the positive value for t turns out to be approximately 3.31 seconds. So it takes around 3.31 seconds for the stone to fall to the water.

Now, let's move on to the speed it strikes the water with. To find this, we can use the equation v = u + at, where v is the final velocity. Plugging in the values:

v = 20 + (-9.8)(3.31)

Calculating the answer... and we find that the speed of the stone striking the water is approximately -11.63 m/s.

So, to sum it all up:
(a) It takes around 3.31 seconds for the stone to fall to the water.
(b) The stone strikes the water with a speed of approximately -11.63 m/s.

To find the time it takes for the stone to fall to the water, we can use the equation of motion for vertical motion in free fall:

h = (1/2)gt^2

where:
h is the vertical distance (53 m in this case)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time

Let's solve for t:

53 = (1/2)(9.8)(t^2)
106 = 9.8t^2
t^2 = 106/9.8
t^2 ≈ 10.82
t ≈ √10.82
t ≈ 3.29 seconds

So it takes approximately 3.29 seconds for the stone to fall to the water.

Now, let's calculate the speed at which the stone strikes the water. We can use the equation for free fall motion:

vf = vi + gt

where:
vf is the final velocity (which is the speed at which it strikes the water)
vi is the initial velocity (20 m/s in this case)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time (3.29 seconds)

Let's calculate vf:

vf = 20 + (9.8)(3.29)
vf ≈ 20 + 32.14
vf ≈ 52.14 m/s

So the stone strikes the water with a speed of approximately 52.14 m/s.