A pilot flying low and slow drops a weight; it takes 2.9 to hit the ground, during which it travels a horizontal distance of 115 . 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?
2.9 seconds. Same amount of gravity.
5 s
To solve this problem, we can use the principles of projectile motion.
Let's break down the given information:
The weight is dropped at a low and slow speed, taking 2.9 seconds to hit the ground. During this time, it travels a horizontal distance of 115 meters.
Now we need to find out how much time it takes for the weight to hit the ground when the pilot does a run at the same height but twice the speed.
Since the run is at twice the speed, we can assume that the initial horizontal velocity of the weight is also doubled.
Using the formula of horizontal distance (d) = initial horizontal velocity (v) × time (t), we can rearrange the equation to find time.
For the initial drop:
115 meters = v1 × 2.9 seconds
We can now solve for v1 (initial horizontal velocity):
v1 = 115 meters / 2.9 seconds
v1 ≈ 39.655 meters/second
Now, for the run at twice the speed:
The weight is dropped with an initial horizontal velocity of 2v1 and time taken is t.
115 meters = (2v1) × t
We can now solve for t:
t = 115 meters / (2v1)
t ≈ 115 meters / (2 × 39.655 meters/second)
t ≈ 1.45 seconds
So, it takes approximately 1.45 seconds for the weight to hit the ground during the run at the same height but twice the speed.