An airplane traveling 1001 meters above the ocean at 35 m/s is to drop a box of supplies to shipwrecked victims below.
a. How many seconds before being directly overhead should the box be dropped?
This is really all the same question
do horizontal d = u t) and vertical (h = (1/2) g t^2) problems separately.
the time, t is the same for each.
see
http://www.jiskha.com/display.cgi?id=1386888436#1386888436.1386888818
To determine how many seconds before being directly overhead the box should be dropped, we need to find the time it takes for the box to fall from the plane to the victims below.
The vertical motion of the box can be described by the equation of motion:
h = ut + (1/2)gt^2
where:
h = vertical displacement (height above the ocean) = -1001 meters (as it is dropping)
u = initial vertical velocity = 0 m/s (as the box is dropped without an initial velocity)
g = acceleration due to gravity = 9.8 m/s^2 (assuming no air resistance)
t = time taken (unknown)
Substituting the given values into the equation, we get:
-1001 = 0t + (1/2)(9.8)t^2
Simplifying the equation gives:
-1001 = 4.9t^2
Dividing both sides by 4.9, we have:
t^2 = -1001 / 4.9
Taking the square root of both sides gives:
t ≈ √( -1001 / 4.9)
Calculating this expression, we get:
t ≈ √(-1001 / 4.9) ≈ √(-204.49) ≈ ± 14.29
Since time cannot be negative in this context, we consider the positive value:
t ≈ 14.29 seconds
Therefore, the box should be dropped approximately 14.29 seconds before the plane is directly overhead.
To find out the time required for the box to be dropped, we need to use the equation of motion. The formula for calculating the time of free fall is given by:
t = sqrt(2h/g)
Where:
t = time of fall
h = height of fall
g = acceleration due to gravity (approximately 9.8 m/s^2 on Earth)
Given:
h = 1001 meters
Substituting the given values into the formula, we get:
t = sqrt(2 * 1001 / 9.8)
Simplifying further:
t = sqrt(2002 / 9.8)
t = sqrt(204.0816)
Therefore, t ≈ 14.28 seconds.
So, the box should be dropped approximately 14.28 seconds before the airplane is directly overhead the shipwrecked victims.