We can use the equations of motion to solve this problem.
a) 4.1 seconds
- We need to find the time it takes for the package to land on the raft. The initial vertical velocity is zero, and the acceleration due to gravity is -9.81 m/s^2 (negative because it opposes the upward motion).
We can use the equation:
h = vt + (1/2)at^2
where h is the initial height (88 m), v is the initial vertical velocity (zero), a is the acceleration due to gravity (-9.81 m/s^2), and t is the time.
Plugging in the values, we get:
88 = 0 x t + (1/2)(-9.81)(t^2)
Simplifying and solving for t, we get:
t = sqrt((2 x 88)/9.81)
t ≈ 4.1 seconds
Therefore, the package is in the air for 4.1 seconds.
b) 1.25 km
- We need to find the horizontal distance travelled by the package in 4.1 seconds. The initial horizontal velocity is 78.3 mph (we convert to m/s), and the acceleration is zero (since there is no horizontal force acting on the package).
We can use the equation:
d = vt
where d is the distance, v is the initial horizontal velocity, and t is the time.
Plugging in the values, we get:
d = 78.3 mph x (1.609 km/m)(1/3600 h/s) x 4.1 s
d ≈ 1.25 km
Therefore, the package lands about 1.25 km east of the raft.
c) 97.5 m/s
- We can use the components of velocity to find the final velocity of the package. The vertical velocity is -gt, where g is the acceleration due to gravity and t is the time of flight (4.1 seconds). The horizontal velocity is 78.3 mph (which we convert to m/s).
The final velocity can be found using the Pythagorean theorem:
vf = sqrt(vh^2 + vv^2)
where vh is the horizontal velocity and vv is the vertical velocity.
Plugging in the values, we get:
vf = sqrt((78.3 mph x (1.609 km/m)(1/3600 h/s))^2 + (-9.81 m/s^2 x 4.1 s)^2)
vf ≈ 97.5 m/s
Therefore, the final velocity of the package is about 97.5 m/s at an angle of tan^-1(-(9.81 m/s^2 x 4.1 s) / (78.3 mph x (1.609 km/m)(1/3600 h/s))) = -0.134 rad = -7.7 degrees below the horizontal.