The Bellagio is about 150 meters tall. A person drops a penny off the roof. The penny is 1 kg. How fast will it be going when it hits the ground? [Type you answer here]

Supreme Scream is about 90 meters tall. You get blasted down at 80 km/h. Use a mass of 1 kg. What will your velocity be after falling 45 meters?
[Type you answer here]

You go sledding down a very hill close to a major highway (don’t ever do this. It is a bad idea…). At the bottom of the hill is a smaller hill (5 meters tall) acting as a barrier to the highway. You get a running start of 4 m/s. Use a mass of 1 kg. How high up the hill can you start without going over the smaller hill into traffic?
[Type you answer here]

1. The final velocity of the penny before it hits the ground is approximately 54.25 m/s

2. The velocity after falling 45 meters is approximately 37.10 m/s

3. The height up the hill one can start without going over the smaller hill is approximately 2.75 meters

Explanation:

The height of the Bellagio, h = 150 meters

The mass of the penny, m  = 1 kg

The kinematic equation of motion that can be used to find the final velocity of the penny 'v' before it hits the ground, is presented as follows;

v² = u² + 2·g·h

Where;

v = The final velocity of the penny after dropping through a height, 'h'

u = The initial velocity of the penny = 0 m/s for the penny initially at rest

g = The acceleration due to gravity ≈ 9.81 m/s²

h = The height from which the penny was dropped = 150 m

∴ v² ≈ 0² + 2 × 9.81 × 150 = 2,943

v ≈ √2,943 ≈ 54.25

The final velocity of the penny before it hits the ground, v ≈ 54.25 m/s

2. Here, the initial velocity, u = 80 km/h = 80 km/h × 1000 m/km × 1 h/(60 × 60 s) = 200/9 m/s = 22. m/s

The height of supreme scream, = 90 meters

The height at which the velocity is required, h = 45 meters

From v² = u² + 2·g·h, we get;

v² = 22.² + 2 × 9.81 × 45 ≈ 1,376.73

∴ v = √1,376.73 ≈ 37.10

The velocity 'v' after falling 45 meters is, v = 37.10 m/s

3. The height of the smaller hill, h = 5 meters

The running start = 4 m/s = The initial velocity

The velocity required to reach the height, h, of the smaller heal v = √(2·g·h)

∴ v = √(2 × 9.81 m/s² × 5 m) ≈ 9.9 m/s

The height 'h'' up the larger hill that will give a velocity, 'v', at the bottom of the smaller hill of approximately 9.9 m/s with an initial velocity, u = 4 m/s, is given as follows;

v² = u² + 2·g·h'

9.9² = 4² + 2 × 9.81 × h'

∴ h' = 9.9²/(4² + 2 × 9.81) ≈ 2.75

Given that the running start is 40 m/s, the height up the hill one can start without going over the smaller hill, h' ≈ 2.75 meters