Answer:
a.) a = 0 ms⁻²
b.) a = 9.58 ms⁻²
c.) a = 7.67 ms⁻²
Explanation:
a.)
Acceleration (a) is defined as the time rate of change of velocity
Given data
Final velocity = v₂ = 0 m/s
Initial velocity = v ₁ = 0 m/s
As the space shuttle remain at rest for the first 2 minutes i.e there is no change in velocity so,
a = 0 ms⁻²
b.)
Given data
As the space shuttle start from rest, So initial velocity is zero
Initial velocity = v₁ = 0 ms⁻¹
Final velocity = v₂ = 4600 ms⁻¹
Time = t = 8 min = 480 s
By the definition of Acceleration (a)

a = 9.58 ms⁻²
c.)
Given data
As the space shuttle is at rest for first 2 min then start moving, So initial velocity is zero
Initial velocity = v₁ = 0 ms⁻¹
Final velocity = v₂ = 4600 ms⁻¹
Time = t = 10 min = 600 s
By the definition of Acceleration (a)

a = 7.67 ms⁻²
Answer:
9.6 Ns
Explanation:
Note: From newton's second law of motion,
Impulse = change in momentum
I = m(v-u).................. Equation 1
Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.
Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)
Substitute into equation 1
I = 2.4[2.5-(-1.5)]
I = 2.4(2.5+1.5)
I = 2.4(4)
I = 9.6 Ns
<u>Answer:</u>
Ball will move 92.8125 meter along the cliff in 7.5 seconds.
<u>Explanation:</u>
We have equation of motion ,
, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this case initial velocity = 0 m/s, acceleration = 3.3
, we need to calculate displacement when time = 7.5 seconds.
Substituting

So ball will move 92.8125 meter along the cliff in 7.5 seconds.
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