Answer:
Acceleration = 9 × 10^5 m/s^2 ( deceleration )
Explanation:
From the first equation of motion:
V = u + at
15000 = 30000 + 60a
a = ( 15000-30000)/60
a = 9 × 10^5 m/s^2
To find the force we use the formula,
F = ma , where m is mass and a acceleration
Using the formula,
F = ma
F = 0.42 x 14.8
F = 6.216 N / 6.22 N
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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)