An American Red Cross plane needs to drop emergency supplies to victims of a hurricane in a remote part of the world. The plane
is traveling horizontally at 100.0 m/s at a height of 50.0 m above the ground. What horizontal distance does the package travel before striking the ground?
1 answer:
Refer to the diagram shown below.
Assume that air resistance is ignored.
Note:
The distance, h, of a falling object with initial vertical velocity of zero at time t is
h = (1/2)gt²
where
g = 9.8 m/s²
The initial vertical velocity of the supplies is 0 m/s.
It the time taken for the supplies to reach the ground is t, then
(50 m) = (1/2)*(9.8 m/s²)*(t s)²
Hence obtain
t² = 50/4.9 = 10.2041
t = 3.1944 s
The horizontal distance traveled at a speed of 100 m/s is
d = (100 m/s)*(3.1944 s) = 319.44 m
Answer: 319.4 m (nearest tenth)
Assume that air resistance is ignored.
Note:
The distance, h, of a falling object with initial vertical velocity of zero at time t is
h = (1/2)gt²
where
g = 9.8 m/s²
The initial vertical velocity of the supplies is 0 m/s.
It the time taken for the supplies to reach the ground is t, then
(50 m) = (1/2)*(9.8 m/s²)*(t s)²
Hence obtain
t² = 50/4.9 = 10.2041
t = 3.1944 s
The horizontal distance traveled at a speed of 100 m/s is
d = (100 m/s)*(3.1944 s) = 319.44 m
Answer: 319.4 m (nearest tenth)
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Answer:
The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.
Explanation:
Given;
mass of the object, m = 2 kg
weigh of the object, W = 20 N
tension on the rope, T = 12 N
The acceleration of the object is calculated by applying Newton's second law of motion as follows;
T = F + W
T = ma + W
ma = T - W
(the negative sign indicates deceleration of the object)
The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.