A bomber flying horizontally at 250 m/s releases a bomb 1500m above the ground. At the point of impact, what were the horizontal speed and vertical speed?
Horizontal speed 250m/s
vertical speed^2=2*g*height
To find the horizontal and vertical speeds at the point of impact, we can apply the principles of projectile motion.
First, we need to understand that the horizontal and vertical motions of the bomb are independent of each other. This means that the horizontal speed remains constant throughout the motion, while the vertical speed changes due to the force of gravity.
Let's calculate the time it takes for the bomb to reach the ground:
We know that the initial vertical velocity (u) is 0 m/s because the bomb is released from rest.
The final vertical displacement (s) is -1500 m (negative because it's directed downward).
The acceleration due to gravity (a) is approximately -9.8 m/s² (negative because it acts downward).
Using the formula: s = ut + (1/2)at², we can rearrange it to solve for time (t):
-1500 = 0 * t + (1/2) * (-9.8) * t²
-1500 = -4.9t²
t² = -1500 / -4.9
t² = 306.12
t ≈ √306.12
t ≈ 17.5 seconds
Now that we know it takes approximately 17.5 seconds for the bomb to reach the ground, we can find the horizontal speed.
The horizontal distance traveled by the bomb can be calculated using the formula: distance = speed * time.
distance = 250 m/s * 17.5 s
distance ≈ 4375 m
Therefore, the horizontal speed at the point of impact is approximately 250 m/s, and the vertical speed can be calculated using the formula: speed = acceleration * time.
speed = -9.8 m/s² * 17.5 s
speed ≈ -171.5 m/s
Hence, at the point of impact, the horizontal speed is approximately 250 m/s, and the vertical speed is approximately -171.5 m/s (directed downward).