Question

The velocity of an aeroplane is $600km/h$ which is flying horizontally and at a height of $1960m$. A bomb is released from the aeroplane when it vertically reaches point $A$. At point $B$, the bomb is struck. What is the distance between $AB$?

Answer 1

Step 1: Given data

The horizontal velocity of an airplane, ${\mathrm{v}}_{\mathrm{x}}=600\mathrm{km}/\mathrm{h}=166.67\mathrm{m}/\mathrm{s}$

Height is $\mathrm{h}=1960\mathrm{m}$

Step 2: Formula used

$h=v_{y}t+\frac{1}{2}g{t}^2$

Here,

h is the height.

t is the time taken.

${\mathrm{v}}_{\mathrm{y}}$ is the vertical velocity.

Step 3: Calculation of distance between $AB$

$h=v_{y}t+\frac{1}{2}g{t}^2$

$1960=0+\frac{1}{2}\times 9.8\times t^2$

$1960=4.9t^2$

$t^2=\frac{1960}{4.9}=400$

$t=\sqrt{400}=20\mathrm{s}$

$AB=v_x\times t$

$=166.67\times 20$

$=3333.4\mathrm{m}$

Therefore, the distance between $AB$ is $333.4m$.