Question

An aeroplane is flying horizontal with a velocity 600km/h and at a height of 1960m when it is vertically at a poin A on the ground a bomb is released from it the bomb strikes the ground at point B the distancs AB will be

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{AB=3.33 km}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about an aeroplane flying horizontally with velocity 600km/h at the height of 1960m.

• Bomb is droped from the aeroplane so initial velocity of bomb is equal to the velocity of aeroplane in horizontal direction.

• We have to find the distance from aeroplane to ground where bomb strikes the ground.

$\green{\underline \bold{Given : }} \\ :\implies \text{velocity \: of \:aeroplane }v_{x} = 600km/hr = 166.67m/s \\\\ :\implies \text{velocity \: of \: plane = velocity \: of \: bomb} \\\\ :\implies \text{initial \: vertical \: velocity}( v_{y}) = 0 \\\\ :\implies \text{height(h) = 1960m }\\\\ :\implies \text{let \: time \: taken = t }\\\\ :\implies a = 9.8m/{s}^{2} \\ \\ \red{\underline \bold{To \: Find : }} \\\\ :\implies \text{distance \: AB = ?}$

• According to given question :• By using formula :

$:\implies s = ut + \frac{1}{2} a {t}^{2} \\\\ :\implies h = v_{y} \times t + \frac{1}{2} \times 9.8 \times {t}^{2} \\\\ :\implies 1960= 0 \times t + 4.9 {t}^{2} \\\\ :\implies {t}^{2} = \frac{1960}{4.9} \\\\ :\implies t = \sqrt{400} \\\\ :\implies t = 20 \: sec$

• We find the time taken to reach ground.• So, acceleration of bomb is zero.

$:\implies AB = v_{x}\times t = 166.7 \times 20 \\\\ \green{:\implies AB = 3333.4 \: m = 3.33 \: km}$

Answer 2

Answer:

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Explanation:

see in the question its given:

Velocity= 600Km/hr = 600x1000/3600 m/s=166.66m/s

Ht=1960m.

To Find: The range of the projectile (R).

Its actually a Projetile motion .

For a projectile Ht=1/2*g*t*t.There fore we get t= 20s.

R=V*t=166.66*20=3333.33m/s=3.333Km/hr.