There are competitions in which pilots fly small planes low over the ground and drop weights, trying to hit a target
A pilot flying low and slow drops a weight; it takes 1.6 s to hit the ground, during which it travels a horizontal distance of 160 m . Now the pilot does a run at the same height but twice the speed. How much time does it take the weight to hit the ground?
See previous post: Fri, 9-11-15, 12:45 PM.
To solve this problem, we can use the concept of time of flight in projectile motion. The time of flight is the time it takes for an object to travel from its initial position to the position where it hits the ground.
Given that the weight takes 1.6 seconds to hit the ground when flying at a certain speed, we can assume that the initial vertical velocity is zero (since there is no mention of upward or downward motion).
Let's use the following variables:
- t1: time taken by the weight to hit the ground when flying at a certain speed (1.6 seconds in this case)
- t2: time taken by the weight to hit the ground when flying at twice that speed (what we need to find)
- v1: initial horizontal velocity when flying at a certain speed
- v2: initial horizontal velocity when flying at twice that speed
- d: horizontal distance traveled by the weight (160 meters in this case)
- g: acceleration due to gravity (approximately 9.8 m/s²)
We can use the equation d = v1 * t1 to find the initial horizontal velocity v1. Rearranging the equation gives v1 = d / t1.
Now, let's find v2 using the relation v2 = 2 * v1, since the speed is twice that of the initial speed.
Substituting the values into the equation gives:
v2 = 2 * (d / t1) = (2 * d) / t1
To find t2, we can use the equation d = v2 * t2 and rearrange it as t2 = d / v2.
Now, let's substitute the values:
t2 = d / v2 = d / [(2 * d) / t1] = t1 / 2
Therefore, the weight will take half the time to hit the ground when flying at twice the speed.
In this case, t2 = t1 / 2 = 1.6 s / 2 = 0.8 seconds.
So, when flying at twice the speed, it will take the weight 0.8 seconds to hit the ground.
To answer this question, we can use the principles of projectile motion. When an object is dropped from rest, its vertical motion can be described by the equations of motion. In this case, the weight is dropped vertically and travels a horizontal distance of 160 m in 1.6 s.
Let's first find the initial vertical velocity of the weight. Since the weight is dropped from rest, its initial vertical velocity is 0 m/s.
Next, we need to find the time it takes for the weight to hit the ground when the pilot is flying at twice the speed. We know that the horizontal distance traveled by the weight is still 160 m, but the pilot is now flying at twice the speed.
If the initial horizontal velocity of the weight in the original scenario is V1, then the initial horizontal velocity in the new scenario will be 2V1 (twice the original velocity).
Now, the time taken for the weight to hit the ground can be found using the equation:
time = horizontal distance / horizontal velocity
In the original scenario, the time is 1.6 s and the horizontal distance is 160 m. Hence, we have:
1.6 s = 160 m / V1
In the new scenario, the time taken will be denoted by t and the horizontal distance is still 160 m. However, the horizontal velocity has doubled, so we have:
t = 160 m / (2V1)
Since the horizontal velocity appears in both equations, we can express it in terms of time by rearranging the first equation:
V1 = 160 m / 1.6 s
By substituting this value into the second equation, we can solve for t:
t = 160 m / (2 * (160 m / 1.6 s))
t = 1.6 s
Therefore, when the pilot does a run at twice the speed, it will still take the weight 1.6 seconds to hit the ground.