An aeroplane flying horizontally at an altitude of 490 m with a speed of 180 kph drops a bomb.the horizontal distance at which it hits the ground is
(A)500m
(B)1000m
(C)250m
(D)50m
Please answer with explination
pahiroy1221:
shymasneha are u sure that these are the only options?........just asking
in option (C) it is 250 m i think it should be 2450 m ?
sorry i was wrong all the options are correct :)
and the answer is option is (A)
Answers
Answered by
126
given,
speed of the airplane = 18 kph = 50 m/sec
it's altitude = 490 meter
1st of all we have to find out the time to travel 490 meter ,
now,
s = ut + 1/2 at²
490 = 0 * t + 1/2 * 9.8 * t²
490 = +4.9t²
490/4.9 = t²
100 = t²
10 = t
therefore time time taken to travel 490 meter = 10 sec
therefore now ,
horizontal distance = time taken * speed
distance = 10* 50
distance = 500 meter .
therefore the answer is 500 meter and the correct option is option(A) that is 500 m .
i hope it helps........................^_^
speed of the airplane = 18 kph = 50 m/sec
it's altitude = 490 meter
1st of all we have to find out the time to travel 490 meter ,
now,
s = ut + 1/2 at²
490 = 0 * t + 1/2 * 9.8 * t²
490 = +4.9t²
490/4.9 = t²
100 = t²
10 = t
therefore time time taken to travel 490 meter = 10 sec
therefore now ,
horizontal distance = time taken * speed
distance = 10* 50
distance = 500 meter .
therefore the answer is 500 meter and the correct option is option(A) that is 500 m .
i hope it helps........................^_^
Answered by
15
Answer:
500m
Explanation:
the speed of aeroplane is 180km/h = 180*5/18 = 50m/s and h = 490m
now , speed of bomb just after the drop is 50m/s horz.
and time of flight is
h = 1/2*(9.8)*(t²)
=> t = √2*490/9.8 = 10sec
so, the distance travelled horizontally for 10 sec is = 50*10 = 500m
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