An airplane is flying with a velocity of 600 km/hr and at a height of 1960 m. When it is vertically above a point A on a ground, a bomb is released frm it. The bomb strikes the ground at pt B. The distance AB is ____ ?
time to fall:
1969=1/2 g t^2
t= sqrt(2*1980/g)
now, horizontal distance traveled during that time t
AB= 600km/hr*t(inseconds)*1/3600 in meters.
To find the distance AB, we need to consider the time it takes for the bomb to hit the ground after it is released.
First, let's calculate the time it takes for the bomb to hit the ground. We know that the bomb is released vertically, so its motion can be analyzed separately from the horizontal motion of the airplane.
The bomb falls freely under the effect of gravity. We can use the formula for the time of flight of an object undergoing free fall:
t = sqrt((2 * h) / g)
Where:
t is the time of flight
h is the initial height (1960 m in this case)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Let's calculate the time:
t = sqrt((2 * 1960) / 9.8)
t ≈ sqrt(400) ≈ 20 seconds
Now that we know the time of flight, we can find the horizontal distance traveled by the airplane during this time. Since the airplane is moving horizontally with a constant velocity, the distance traveled is given by:
distance = velocity * time
distance = 600 km/hr * (20 s / 3600 s) [convert km/hr to m/s]
distance = (600 * 1000 m/3600 s) * 20 s
distance = (600 * 1000 * 20) / 3600
distance ≈ 3333.33 m
Therefore, the distance AB is approximately 3333.33 meters.