A bomber drops a bomb from an aircraft flying at 2000Km/h with an altitude of 7 Km. The speed at which the bomb will strike the earth would be.
a) Greater than 2000Km/h
b) Less than 2000 Km/h
c) Equal to 2000 Km/h
d) Can not compare with the given data
V^2 = Vo^2 + 2g*h = 0 + 19.6*7000, V = 370 m/s.
V = 0.370km/s * 3600s/h = 1333 km/h.
To determine the speed at which the bomb will strike the Earth, we need to consider its horizontal and vertical motion separately.
First, let's consider the horizontal motion. Since the bomber is flying at a constant speed of 2000 km/h, the horizontal velocity of the bomb will also be 2000 km/h when it is dropped. This means that the horizontal speed at which the bomb strikes the Earth will be 2000 km/h.
Now, let's consider the vertical motion. The bomb is dropped from an altitude of 7 km, so it will undergo free fall due to gravity. The acceleration due to gravity is approximately 9.8 m/s^2. To compare this with the horizontal speed, we need to convert the altitude from kilometers to meters. 1 km is equal to 1000 m, so 7 km is equal to 7000 m.
We can use the kinematic equation for free fall:
vf^2 = vi^2 + 2ad
where vf is the final vertical velocity, vi is the initial vertical velocity (which is 0 since the bomb is dropped), a is the acceleration, and d is the distance.
Using this equation, we can find the vertical velocity at which the bomb strikes the Earth:
vf^2 = 0 + 2 * 9.8 m/s^2 * 7000 m
vf^2 = 2 * 9.8 m/s^2 * 7000 m
vf^2 = 137200 m^2/s^2
Taking the square root of both sides, we find:
vf = √(137200 m^2/s^2)
vf ≈ 370.9 m/s
Since we are asked for the speed in km/h, we can convert the vertical speed from m/s to km/h:
vf ≈ 370.9 m/s * (3600 s/1 hr) * (1 km/1000 m)
vf ≈ 1331 km/h
Therefore, the speed at which the bomb will strike the Earth is approximately 1331 km/h.
So the correct answer is:
b) Less than 2000 Km/h.