Constant Acceleration - 3

The ball is dropped, so u = 0

Using v² = u² + 2gh

v² = 0 + 2 × 10 × 125

v² = 2500

v = √2500

v = 50 m/s

At maximum height, final velocity v = 0

Using v² = u² − 2gH

0 = u² − 2gH

H = u² / 2g

H = (50)² / (2 × 10)

H = 125 m

Using v² = u² + 2gh

v² = (10)² + 2 × 10 × 125

v² = 2600

v = 50.990195135928 m/s

Since the ball is dropped, u = 0

Using v² = u² + 2gh

h = (v² − u²) / 2g

h = (50² − 0) / (2 × 10)

h = 125 m

Using v² = u² + 2gh

h = (v² − u²) / 2g

h = (50² − 10²) / (2 × 10)

h = 2400 / 20

h = 120 m

Since the car starts from rest, u = 0

Using v² = u² + 2as

s = (v² − u²) / 2a

s = (100² − 0) / (2 × 10)

s = 500 m

For minimum stopping distance, the car must decelerate at its maximum rate

Initial velocity, u = 100 m/s

Final velocity, v = 0

Acceleration, a = −25 m/s²

Using v² = u² + 2as

0 = (100)² + 2(−25)s

s = 200 m

The car starts from rest, so u = 0

Convert 100 km/h into m/s

v = 100 × 5/18 = 27.78 m/s

Using v² = u² + 2as

s = (v² − u²) / 2a

s = (27.78)² / (2 × 12)

s = 32.15 m

Initial velocity, u = 100 × 5/18 = 27.78 m/s

Final velocity, v = 0

Acceleration, a = −10 m/s²

Using v² = u² + 2as

0 = (27.78)² + 2(−10)s

s = 38.58 m