A bomb dropped from a aeroplane in level flight hits a target on ground which is 500m ahead of point of dropping. If the speed of aeroplane is 50m/s. Calculate the height at which it is flying.[g=9.8ms^2]
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Answered by
7
Horizontal distance traveled = s = 500 m
u = 50 m/s
g = 9.8 m/s²
The horizontal velocity of the bomb remains constant at u = 50 m/s.
Time of flight = t = s / u = 500 / 50 = 10 sec.
Initial vertical velocity = 0
Height descended in time t = 10 sec.: h
h = u t + 1/2 g t²
= 0 + 1/2 * 9.8 * 10²
h = 490 meters.
The aeroplane was flying at 490 m. above Earth's surface.
u = 50 m/s
g = 9.8 m/s²
The horizontal velocity of the bomb remains constant at u = 50 m/s.
Time of flight = t = s / u = 500 / 50 = 10 sec.
Initial vertical velocity = 0
Height descended in time t = 10 sec.: h
h = u t + 1/2 g t²
= 0 + 1/2 * 9.8 * 10²
h = 490 meters.
The aeroplane was flying at 490 m. above Earth's surface.
Answered by
1
Answer:
h=490m
Explanation:
s=500m
u=50m/s
g=9.8ms^2
u=50
t=s/u
=> 500/50 =10
height decreased in time =10s:h
h=ut+1/2*gt^2
= 0+1/2*9.8*10^2
h=490m
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