Question

a bomb dropped from an aeroplane when it was at a height of 196m above point A on the ground and was moving horizontally at a speed of 100m/s.find the time and distance from A to the point where the bomb strikes the ground​

Answer 1

Answer :

• time = 2√10 s
• distance = 632 m

Solution :

Along y - direction :

• height of the aeroplane from the aeroplane from the ground = 196 m

• acceleration due to gravity = 9.8 m /s²

• As the bomb is dropped the initial velocity of the bomb = 0 m /s

Let's use the second equation of motion ,

$\bigstar \boxed{ \mathtt{h = ut + \dfrac{1}{2} a {t}^{2} }}$

here ,

• h = height
• u = initial velocity
• a = acceleration
• t = time taken

Now substituting the given values in the above equation we get ,

$\rightarrow \mathtt{196 = 0 \times t + \dfrac{1}{2} \times 9.8 \times {t}^{2} }$

$\rightarrow \mathtt{196 = 4.9 \times {t}^{2} }$

$\rightarrow \mathtt{ {t}^{2} = \dfrac{196}{4.9} }$

$\rightarrow \mathtt{t = \sqrt{40} }$

$\rightarrow \orange{\mathtt{t = 2 \sqrt{10} \: s}}$

______________________

Along the x - direction :

height of the aeroplane from the aeroplane from the ground = 196 m

velocity along the horizontal = 100 m /s

acceleration = 0 m/s²

Let's use the second equation of motion ,

$\bigstar \boxed{ \mathtt{s = ut + \dfrac{1}{2} a {t}^{2} }}$

here ,

• h = height
• u = initial velocity
• a = acceleration
• t = time taken

Now substituting the given values in the above equation we get ,

$\rightarrow \mathtt{s = 100 \times 2\sqrt{10} + \dfrac{1}{2} \times 0 \times 40}$

$\rightarrow \mathtt{s = 100 \times 2\sqrt{10} }$

$\rightarrow \mathtt{s = 100 \times 2 \times 3.16 }$

$\rightarrow \mathtt{s= 100 \times \: 6.32}$

$\rightarrow \red{\mathtt{s = 632 \: m}}$

Answer 2

$\huge\underline{\red{A}\pink{N}\blue{S}\purple{W}\orange{E}\blue{R}}$

• time = 2√10 s
• distance = 632 m

Solution

Along y - direction :

height of the aeroplane from the aeroplane from the ground = 196 m

acceleration due to gravity = 9.8 m /s²

As the bomb is dropped the initial velocity of the bomb = 0 m /s

Let's use the second equation of motion ,

$\star\boxed{h=ut+\frac{1}{2}at^2}$

here ,

h = height

u = initial velocity

a = acceleration

t = time taken

Now substituting the given values in the above equation we get ,

$196=0×t+ \frac{1}{2}×9.8×t^2$

$\mathtt{196 = 4.9 \times {t}^{2} }$

$\mathtt{ {t}^{2} = \dfrac{196}{4.9} }$

$\mathtt{t = \sqrt{40} }$

$\pink{\mathtt{t = 2 \sqrt{10} \: s}}$

_________________________

Along the x - direction :

height of the aeroplane from the aeroplane from the ground = 196 m

velocity along the horizontal = 100 m /s

acceleration = 0 m/s²

Let's use the second equation of motion ,

$\boxed\star{s=ut+\frac{1}{2}at^2}$

here ,

h = height

u = initial velocity

a = acceleration

t = time taken

Now substituting the given values in the above equation we get ,

$\mathtt{s = 100 \times 2\sqrt{10} + \dfrac{1}{2} \times 0 \times 40}$

$s=100×2\sqrt{10}$

$arrows=100×2×3.16$

$\mathtt{s= 100 \times \: 6.32}$

$\red{\mathtt{s = 632 \: m}}$

HOPE IT HELPS!!