Question

An aeroplane is flying horizontally with a velocity of 600 km/h and at a height of 1960 m. When it is vertically at a point A on the ground a bomb is released from it. The bomb strikes the ground at point B. The distance AB is

Answer 1

When the bomb is dropped, it will have an initial horizontal velocity which is equal to the speed of the aeroplane. So the bomb fall and travel forward too.

initial horizontal velocity,$v_x$ = 600 km/h
$v_x=600\ km/h = 600* \frac{5}{18}\ m/s=166.67\ m/s$
initial vertical velocity,$v_y$ = 0
height, h = 1960m
time taken to fall = t

$h=v_yt+ \frac{1}{2}gt^2\\ \\1960=0+ \frac{1}{2}*9.8*t^2\\ \\1960=4.9t^2\\ \\t^2= \frac{1960}{4.9}=400\\ \\t= \sqrt{400}=20s$

In 20s, the horizontal distance travelled is AB

$AB=v_x*t=166.67*20=\boxed{3333.4\ m}$

Answer 2

Answer:

When the bomb is dropped, it will have an initial horizontal velocity which is equal to the speed of the aeroplane. So the bomb fall and travel forward too.

initial horizontal velocity,v_xv

x

= 600 km/h

v_x=600\ km/h = 600* \frac{5}{18}\ m/s=166.67\ m/sv

x

=600 km/h=600∗

18

5

m/s=166.67 m/s

initial vertical velocity,v_yv

y

= 0

height, h = 1960m

time taken to fall = t

\begin{lgathered}h=v_yt+ \frac{1}{2}gt^2\\ \\1960=0+ \frac{1}{2}*9.8*t^2\\ \\1960=4.9t^2\\ \\t^2= \frac{1960}{4.9}=400\\ \\t= \sqrt{400}=20s\end{lgathered}

h=v

y

t+

2

1

gt

2

1960=0+

2

1

∗9.8∗t

2

1960=4.9t

2

t

2

=

4.9

1960

=400

t=

400

=20s

In 20s, the horizontal distance travelled is AB

AB=v_x*t=166.67*20=\boxed{3333.4\ m}AB=v

x

∗t=166.67∗20=

3333.4 m