A bomb is dropped from a plane flying at constant horizontal velocity at an altitude of 600m. Where is the bomb after 2 sec? 5 sec? 10 sec? what are the velocities said period?
h(t) = ho - 0.5g.t^2
h(2) = 600 - 4.9*2^2 = 580.4 m. Above gnd.
V(2) = Vo + g*t = 0 + 9.8*2 = 19.6 m/s.
h(5) = 600 - 4.9*5^2 = 477.5 m. Above gnd.
V(5) = 0 + 9.8*5 = 49 m/s.
h(10) = 600 - 4.9*100 = 110 m. Above gnd.
V(10) = 0 + 9.8*10 = 98 m/s.
To determine the position and velocity of the bomb at various time intervals, we can use the equations of motion. Assuming air resistance is negligible, the horizontal velocity remains constant, while the vertical position and velocity are affected by gravity.
Given:
Initial altitude (h) = 600 m
Time intervals: t1 = 2 sec, t2 = 5 sec, t3 = 10 sec
Horizontal velocity (v) = constant
1. Position after 2 seconds:
Using the formula for vertical position:
h = h0 + v0t + (1/2)gt^2
Since the initial vertical velocity (v0) is 0 m/s (dropped vertically), the equation simplifies to:
h = h0 + (1/2)gt^2
Substituting the values:
h = 600 + (1/2)(9.8)(2)^2
h = 600 + 19.6
h = 619.6 m
Therefore, after 2 seconds, the bomb is at an altitude of 619.6 meters.
2. Position after 5 seconds:
Using the same formula:
h = h0 + (1/2)gt^2
Substituting the values:
h = 600 + (1/2)(9.8)(5)^2
h = 600 + 122.5
h = 722.5 m
After 5 seconds, the bomb is at an altitude of 722.5 meters.
3. Position after 10 seconds:
Using the formula:
h = h0 + (1/2)gt^2
Substituting the values:
h = 600 + (1/2)(9.8)(10)^2
h = 600 + 490
h = 1090 m
After 10 seconds, the bomb is at an altitude of 1090 meters.
Since the horizontal velocity remains constant, it does not change throughout the motion, so the velocity at each time interval will be the same.
To find the position of the bomb at different times and the corresponding velocities, we need to use the kinematic equations of motion. Here's how you can solve this problem step by step:
1. Define the given values:
- Initial altitude (y₀) = 600m
- Time elapsed (t) = 2s, 5s, 10s
- Initial velocity in the horizontal direction (v₀x) = constant
2. Calculate the vertical position (y) of the bomb at each given time using the equation of motion for vertical motion:
- y = y₀ + v₀yt - 1/2gt²
- Since the plane is flying horizontally, the initial vertical velocity (v₀y) is 0.
- The acceleration due to gravity (g) is approximately 9.8m/s².
For 2s:
y = 600m + 0m/s * 2s - 1/2 * 9.8m/s² * (2s)²
For 5s:
y = 600m + 0m/s * 5s - 1/2 * 9.8m/s² * (5s)²
For 10s:
y = 600m + 0m/s * 10s - 1/2 * 9.8m/s² * (10s)²
3. Calculate the horizontal position (x) of the bomb at each given time. Since the plane is flying at a constant horizontal velocity, the horizontal displacement will be given by:
- x = v₀x * t
4. Calculate the horizontal velocity (vx) of the bomb. It remains constant throughout the motion, assuming no other forces are acting on it.
Now, let's calculate the results:
For 2s:
- y = 600m - 19.6m = 580.4m (the sign changes because the bomb is falling downward)
- x = v₀x * 2s
- vx = v₀x (since the horizontal velocity remains constant)
For 5s:
- y = 600m - 122.5m = 477.5m
- x = v₀x * 5s
- vx = v₀x
For 10s:
- y = 600m - 490m = 110m
- x = v₀x * 10s
- vx = v₀x
Note: The horizontal velocity (vx) remains constant throughout the motion because there are no horizontal forces acting on the bomb.