Question

A bomb is dropped feom an aeroplane flyinghorizontally with a belocity 720 km/hr at an altitude of 980m. The bomb will hit the ground after a time.

Answer 1

Explanation:

Please refer to the above attachment for the answer.....

Answer 2

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

${\bold{\therefore Time\:taken=14\:sec}}$

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

• In the given question information given about a bomb is dropped feom an aeroplane flyinghorizontally with a belocity 720 km/hr at an altitude of 980m.

• We have to find the time after which bomb hit the ground.

$\underline \bold{Given : } \\ \implies Initial \: velocity( v_{x }) = 720 \: km/h \\ \\ \implies Height(s) = 980 \: m \\ \\ \implies Acceleration(a) = 10 \: m /{s}^{2} \\ \\ \underline \bold{To \: Find : } \\ \implies Time \: taken = ?$

• According to given question :

$\bold{ v_{x} = 720 \times \frac{5}{18} = 200 \: m/s} \\ \\ \bold{Initial \: vertical \: velocity(v_{y} )= 0 \: m/s} \\ \implies h = v_{y}t + \frac{1}{2} g {t}^{2} \\ \\ \implies 980 = 0 + \frac{1}{2} \times 10 \times {t}^{2} \\ \\ \implies {t}^{2} = \frac{980 \times 2}{10} \\ \\ \implies {t}^{2} = 196 \\ \\ \implies t = \sqrt{196} \\ \\ \bold{\implies t = 14 \: sec}$