Can someone help me to understand how to work out this problem?
The pilot of an airplane traveling 240 km/h wants to drop supplies to flood victims isolated on a patch of land 160 m below. The supplies should be dropped how many seconds before the plane is directly overhead?
how long does it take an object to fall 160 m?
Bobpursley -- I do not understand. The answer was 5.7 s and I have no idea what equation was used to get that.
y = .5gt^2
160 = .5gt^2
= 5.71s
To solve this problem, we need to find the time it takes for the supplies to reach the ground after they are dropped from the plane. We can then subtract this time from the time it takes for the plane to travel directly over the patch of land.
Let's start by finding the time it takes for the supplies to fall to the ground. We can use the equation for the distance traveled by a freely falling object:
distance = (1/2) * acceleration * time^2
In this case, the distance is 160 m and the acceleration is due to gravity, which is approximately 9.8 m/s^2. We can rearrange the equation to solve for time:
time^2 = (2 * distance) / acceleration
time = sqrt((2 * distance) / acceleration)
Substituting the values, we get:
time = sqrt((2 * 160 m) / 9.8 m/s^2)
time = sqrt(320 / 9.8) s
time ≈ 5.048 s (rounded to 3 decimal places)
Now we need to find the time it takes for the plane to travel directly over the patch of land. We can use the formula:
time = distance / speed
In this case, the distance is 0 km (since the plane is directly overhead) and the speed is 240 km/h. We need to convert the speed from km/h to m/s:
speed = 240 km/h * (1000 m/km) / (3600 s/h)
speed ≈ 66.67 m/s (rounded to 2 decimal places)
Now we can find the time it takes for the plane to travel directly overhead:
time = distance / speed
time = 0 / 66.67 s
time = 0 s
Finally, we subtract the time it takes for the supplies to reach the ground from the time it takes for the plane to travel directly overhead:
time difference = time - time
time difference = 0 s - 5.048 s
time difference ≈ -5.048 s (rounded to 3 decimal places)
So, the supplies should be dropped approximately 5.048 seconds before the plane is directly overhead.