A plane is flying horizontally at a speed of 290m/s. He drops a package (ignoring air resistance).
A)If it take 15s for the package to hit the ground, how high above is the plane?
B) How far (horizontally) before the island should Steven drop the package?
C) Assuming the airplane does not change speed or elevation, where is the airplane when the package hits the ground?
(If you can't answer these questions just give the equations for I can ATTEMPT them.. Thank you)
A. h = 0.5g*t^2 = 4.9*15^2 = 1103 m.
B. d = 290m/s * 15s = 4350 m.
C. Over the island.
To answer these questions, we can use the equations of motion. Here's how you can solve each part:
A) To find the height above the plane, we can use the equation of motion for vertical motion:
h = ut + (1/2)gt^2
where:
h = height above the ground
u = initial vertical velocity (0 in this case, as the package is dropped)
g = acceleration due to gravity (9.8 m/s^2)
t = time taken for the package to hit the ground (15 s)
Plugging in the values, we get:
h = 0 * 15 + (1/2) * 9.8 * (15^2)
Simplifying the equation gives us the answer for part A.
B) To find how far horizontally before the island Steven should drop the package, we need to consider the horizontal velocity of the plane. Since the plane is flying horizontally, the horizontal velocity will remain constant. We can use the equation:
d = v * t
where:
d = horizontal distance
v = horizontal velocity (290 m/s)
t = time taken for the package to hit the ground (15 s)
Plugging in the values, we get:
d = 290 * 15
Simplifying the equation gives us the answer for part B.
C) To find the position of the airplane when the package hits the ground, we need to consider both the horizontal and vertical positions. Since the horizontal velocity of the plane is constant, the plane will be at the same horizontal position throughout. For the vertical position, we can use the equation:
y = ut + (1/2)gt^2
where:
y = vertical position (height above the ground)
u = initial vertical velocity (0 in this case, as the package is dropped)
g = acceleration due to gravity (9.8 m/s^2)
t = time taken for the package to hit the ground (15 s)
Plugging in the values for y, u, g, and t, we can find the vertical position of the airplane. However, since the initial vertical velocity is zero, the vertical position will also be zero.
So, the answer for part C is that the airplane is at a height of zero when the package hits the ground.
I hope this explanation helps you in solving these questions!