A golf ball rolls off a horizontal cliff with an initial speed of 12.2 m/s. The ball falls a vertical distance of 17.3 m into a lake below. (a) How much time does the ball spend in the air? (b) What is the speed v of the ball just before it strikes the water?

(a) How much time does the ball spend in the air?

Well, it seems like the ball has made quite the ambitious leap into the lake. To calculate the time the ball spends in the air, we can use a little physics called projectile motion.

The ball is rolling off a horizontal cliff, and we are given its initial speed as 12.2 m/s. Since there are no horizontal forces acting on the ball, its horizontal speed will remain constant throughout its flight. This means that we can focus on the vertical motion.

To find the time spent in the air, we can use the equation:

Time = (2 * vertical distance) / g

Where g is the acceleration due to gravity. Since the ball is falling downwards, we take the value of g to be positive.

Using the given vertical distance of 17.3 m, we can plug it into the equation:

Time = (2 * 17.3) / g

Now, using my trusty calculator, I can tell you that the approximate value of g is 9.8 m/s². Plugging in this value, we get:

Time = (2 * 17.3) / 9.8

Solving the equation, we find:

Time ≈ 3.53 seconds

So, the ball spends approximately 3.53 seconds in the air. Hopefully, it doesn't decide to take a detour!

(b) What is the speed v of the ball just before it strikes the water?

Ah, the grand finale! To find the speed of the ball just before it strikes the water, we need to use a little bit of basic kinematics.

Given the vertical distance the ball falls, we can use the equation:

v² = u² + 2 * a * s

Where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the vertical distance fallen.

The initial velocity of the ball is 12.2 m/s, the acceleration due to gravity is approximately 9.8 m/s² (since it's falling downwards), and the vertical distance is 17.3 m.

Plugging these values into the equation, we get:

v² = (12.2)² + 2 * 9.8 * 17.3

v² ≈ 148.84 + 338.68

v² ≈ 487.52

Taking the square root of both sides, we get:

v ≈ √487.52

v ≈ 22.06 m/s

So, just before it strikes the water, the speed of the ball is approximately 22.06 m/s. That's quite a splash!

To solve this problem, we can use the equations of motion for projectile motion.

(a) The time spent in the air can be found using the equation:

Δh = Vyi * t + (1/2) * g * t^2

Where:
Δh = vertical distance (17.3 m)
Vyi = initial vertical velocity (0 m/s, as the ball is rolling horizontally)
g = acceleration due to gravity (-9.8 m/s^2, assuming downward direction)
t = time (unknown)

Rearranging the equation and solving for t:

17.3 = 0 * t + (1/2) * (-9.8) * t^2
17.3 = -4.9 * t^2
t^2 = -17.3 / -4.9
t^2 = 3.530612244897959
t ≈ √3.5306
t ≈ 1.88 seconds

Therefore, the golf ball spends approximately 1.88 seconds in the air.

(b) The final velocity just before the ball strikes the water can be found using the equation:

Vf^2 = Vi^2 + 2 * a * d

Where:
Vf = final velocity (unknown)
Vi = initial velocity (12.2 m/s, horizontal velocity)
a = acceleration (-9.8 m/s^2, vertical acceleration)
d = distance (17.3 m, negative as it is downward)

Rearranging the equation and solving for Vf:

Vf^2 = (12.2)^2 + 2 * (-9.8) * (-17.3)
Vf^2 = 148.84 + 339.64
Vf^2 = 488.48
Vf ≈ √488.48
Vf ≈ 22.10 m/s

Therefore, the speed of the ball just before it strikes the water is approximately 22.10 m/s.

To find the time the ball spends in the air, we can use the kinematic equation for vertical motion. The equation is:

Δy = v₀y * t + (1/2) * g * t²

Where:
Δy is the vertical distance covered (17.3 m),
v₀y is the initial vertical velocity (0 m/s since the ball is rolling off a horizontal cliff),
g is the acceleration due to gravity (approximately 9.8 m/s²),
t is the time.

Since the ball starts at rest in the vertical direction, the first term becomes zero, and we are left with:

Δy = (1/2) * g * t²

Plugging in the known values:

17.3 = (1/2) * 9.8 * t²

Now, we can solve for t²:

t² = (2 * 17.3) / 9.8

t² ≈ 3.52

Taking the square root of both sides gives us the time spent in the air:

t ≈ √3.52

t ≈ 1.88 s

Therefore, the ball spends approximately 1.88 seconds in the air.

To determine the speed of the ball just before it strikes the water, we can use the equation for velocity in vertical motion:

v = v₀y + g * t

Since v₀y is 0 (the ball rolls horizontally), the equation simplifies to:

v = g * t

Plugging in the values:

v = 9.8 * 1.88

v ≈ 18.42 m/s

Therefore, the speed of the ball just before it strikes the water is approximately 18.42 m/s.

1. a) Dy = Viy(t) + (1/2)(g)t^2

15.5 m = 0(t) + (1/2)(9.8 m/s^2)(t^2)
t = sqrt[(15.5 m)/(4.9 m/s^2)]
t = 1.8 sec answer

b) Let V = speed of the ball just before it strikes the water.
Vix = Vfx = 11.4 m/s
Vfy = Viy + gt
Vfy = 0 + (9.8 m/s^2)(1.8 s)
Vfy = 17.64 m/s

V = sqrt[Vfx^2 + Vfy^2]
V = sqrt[(11.4 m/s)^2 + (17.64 m/s)^2]
V = 21.0 m/s

It works I just used different numbers to test.