Question

Standing at the edge of a cliff 32.5m high, you do a ball. Later, you throw a second ball downward with an initial speed of 11m/s. Calculate the velocity if each of the balls when it hits the ground

Answer 1

Answer: The displacement of the ball

ℎ =

2 −

2

2

So, final speed of the first ball

1 = √2ℎ = √2 × 9.8 × 32.5 = 25.24

m

s

The change of it speed

∆1 = 25.24 − 0 = 25.34

m

s

The final speed of the second ball

2 = √

2 + 2ℎ = 27.53

m

s

The change of the second ball speed

∆2 = 27.53 − 11 = 16.53

m

s

Thus

∆1 > ∆2

Answer: ∆1 > ∆2

Explanation:

Answer 2

The final velocity of the first ball is 25.24 m/s and the final velocity of the second ball is 27.53 m/s

To calculate the final velocity of the first ball, we use the equation

v² = u² - 2gh where u = initial velocity of first ball = 0 m/s, v = final velocity of first ball, g = acceleration due to gravity = -9.8 m/s² (since it is directed downwards)and h = height of cliff = 32.5 m

Substituting the values of the variables into the equation, we have

v² = u² - 2gh

v² = (0 m/s)² - 2 × -9.8 m/s² ×  32.5 m

v² = 0 m²/s² + 637 m²/s²

v² = 637 m²/s²

v = √(637 m²/s²)

v = 25.24 m/s

To find the final velocity of the second ball, we also use

v'² = u'² - 2gh where u' = initial velocity of second ball = 11 m/s, v' = final velocity of second ball, g = acceleration due to gravity = -9.8 m/s² (since it is directed downwards)and h = height of cliff = 32.5 m

Substituting the values of the variables into the equation, we have

v'² = u'² - 2gh

v'² = (11 m/s)² - 2 × -9.8 m/s² ×  32.5 m

v'² = 121 m²/s² + 637 m²/s²

v'² = 758 m²/s²

v' = √(758 m²/s²)

v' = 27.53 m/s

So, the final velocity of the first ball is 25.24 m/s and the final velocity of the second ball is 27.53 m/s

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