11 An aeroplane is travelling horizontally at a speed of 80 ms", and drops a crate of emergency supplies
(Figure 2.39). To avoid damage, the maximum vertical speed of the crate on landing is 20 ms.
You may assume air resistance is negligible.
80 ms
Figure 2.39 For End-of-chapter Question 11.
a Calculate the maximum height of the aeroplane when the crate is dropped.
b Calculate the time taken for the crate to reach the ground from this height.
• The aeroplane is travelling at the maximum permitted height Calculate the horizontal distance travelled
by the crate after it is released from the aeroplane.
12
12

Using v² = u² - 2gs where u = initial vertical speed of crate = -20 m/s(since it is falling), v = final vertical speed of crate = 0 m/s (since it stops immediately it lands), g = acceleration due to gravity = 9.8 m/s² and s = maximum height of airplane when crate is dropped.

So, making s subject of the formula, we have

s = -(v² - u²)/2g

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

s = (v² - u²)/2g

s = -((0 m/s)² - (-20 m/s)²)/(2 × 9.8 m/s²)

s = -(-400 m²/s²)/(19.6 m/s²)

s = 400 m²/s²/19.6 m/s²

s = 20.41 m

So, the maximum height of the aeroplane when the crate is dropped is 20.41 m

b.

The time taken for the crate to reach the ground from maximum height is 2.041 s

Using v = u + gt where u = initial vertical speed of crate = -20 m/s(since it is falling), v = final vertical speed of crate = 0 m/s (since it stops immediately it lands), g = acceleration due to gravity = 9.8 m/s² and t = time taken for the crate to reach the ground from maximum height.

Making t subject of the formula, we have

t = (v - u)/g

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

t = (v - u)/g

t = (0 m/s - (-20 m/s))/9.8 m/s²

t = (0 m/s + 20 m/s))/9.8 m/s²

t = 20 m/s/9.8 m/s²

t = 2.041 s

So, the time taken for the crate to reach the ground from maximum height is 2.041 s

c.

The horizontal distance travelled by the crate after it is released from the aeroplane is 163.27 m

Since the horizontal velocity of the crate is the horizontal velocity of the crate, we calculate the horizontal distance the crate travelled after it is released from the aeroplane.from

d = Vt where V = horizontal velocity of crate = 80 m/s and t = time taken for crate to drop from maximum height = 2.041 s.

So, substituting the values of the variables into the equation, we have

d = Vt

d = 80 m/s × 2.041 s

d = 163.27 m

So, the horizontal distance travelled by the crate after it is released from the aeroplane is 163.27 m

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