Question

An aeroplane flying at horizontal at an altitude of 490 metres with a speed of 180 km per hour drops a bomb

Answer 1

Correct question

An aeroplane flying horizontally at an altitude of 490 m with a speed of 180 kmph drops a bomb. The horizontal distance at which it hits the ground is

Given

An aeroplane flying horizontally at an altitude of 490 m with a speed of 180 kmph drops a bomb

To Find

The horizontal distance at which it hits the ground is

Solution

Calculating "Time of flight"

We know that

$\rm \longrightarrow T=\sqrt{\dfrac{2h}{g}}$

Here

• T = Time of flight
• h = Height
• g = acceleration due to gravity

Given that

An aeroplane flying horizontally at an altitude of 490 m with a speed of 180 kmph drops a bomb

Hence

• h = 490 m

We know that

• g = 9.8 m/s²

Substituting the values

We get

$\rm \longrightarrow T=\sqrt{\dfrac{2 \times 490}{9.8}} \ s$

$\rm \longrightarrow T=\sqrt{\dfrac{980}{9.8}} \ s$

$\rm \longrightarrow T=\sqrt{100} \ s$

$\rm \longrightarrow T=10 \ s$

Calculating "Horizontal distance"

We know that

$\rm \longrightarrow R=vT$

Here

• R = Horizontal distance
• v = velocity
• T = Time of flight

According the question

We are asked to find the horizontal distance

Therefore

We must find "R"

Given that

An aeroplane flying horizontally at an altitude of 490 m with a speed of 180 kmph drops a bomb

Hence

• v = 180 kmph = 50 m/s

We found out that

• T = 10 s

Substituting the values

We get

$\rm \longrightarrow R=50 \times 10 \ m$

Therefore

$\rm \longrightarrow R=500 \ m$

Hence

Horizontal distance = 500 m

Answer 2

Given :

• speed of the airplane = 18 kph

• it's altitude = 490 meter 

To Find :

• Find the time and distance

Solution :

$\\ \\ : \implies \: \: \: \: \: \: \boxed{ \sf \:s= ut + \frac{1}{2} a {t}^{2} }$

Substitute all values :

$\\ \\ : \implies \: \: \: \: \: \: \sf \:490= 0 + \frac{1}{ \cancel{2}} \times \cancel{9.8} \times {t}^{2} \\ \\ \\ : \implies \: \: \: \: \: \: \sf \:490 = 4.9 {t}^{2} \\ \\ \\ : \implies \: \: \: \: \: \: \sf \:{t}^{2} = \cancel{\frac{490}{4.9}}t \\ \\ \\ : \implies \: \: \: \: \: \: \sf \: {t}^{2} = 100 \\ \\ \\ : \implies \: \: \: \: \: \: \sf \:t = 10$

$\\ \\ : \implies \: \: \: \: \: \: \sf \:distance = time \times speed$

Substitute all values :

$\\ \\ : \implies \: \: \: \: \: \: \sf \:distance = 10 \times 50 \\ \\ \\ : \implies \: \: \: \: \: \: \sf \:distance = 500$

Hence ,Horizontal distance = 500