Question

an aeroplane is flying horizontally with a velocity of 600km/hr and at a height of 1960 meter. when it is Vertically above a point A on the ground, a bomb is released from it.The bomb strikes the ground at a point But. the distance AB is (ignoring air resistance)​

Answer 1

Answer:

$\huge{\pink{\green{\sf{\therefore AB=3.33 km}}}}$

Explanation:

$\huge{\pink{\green{\underline{\red{\sf{SOLUTION-}}}}}}$

• 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.

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

• 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.

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

Answer 2

Answer:

Explanation:600 km/hr = 500/3 m/s

R = u(2H/g)^1/2

=500/3(2×1960÷9.8)^1/2

=10000/3

=3333.3