An object is dropped out of an airplane that is moving horizontally at 350 m/s and is 22,000 m above the ground. Ignoring fricti
1 answer:
Answer:
Vx = 350m/s
Vy = 656.99 m/s
Explanation:
at the time of drop, the horizontal velocity of the object is same as of plane.
Since friction is ignored, horizontal velocity remains unchanged.
Vx = 350m/s
but the Vy increases due to gravitational acceleration
V^2 - U^2 = 2as
V: FINAL VELOCITY
U: INITIAL VELOCITY
U = 0 m/s
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Answer:
The BMX lands 5.4 m from the end of the ramp.
Explanation:
Hi there!
The position of the BMX is given by the position vector "r":
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
r = position vector at time t
x0 = initial horizontal position
v0 = initial velocity
α = jumping angle
y0 = initial vertical position
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)
Please, see the attached graphic for a better understanding of the situation. At final time, when the bicycle reaches the ground, the vector position will be "r final" (see figure). The y-component of the vector "r final" is - 2.4 m (placing the origin of the frame of reference at the jumping point). With that information, we can use the equation of the y-component of the vector "r" (see above) to calculate the time of flight. With that time, we can then obtain the x-component (rx in the figure) of the vector "r final". Then:
y = y0 + v0 · t · sin α + 1/2 · g · t²
-2.4 m = 0 m + 5.9 m/s · t · sin 40° - 1/2 · 9.8 m/s² · t²
0 = -4.9 m/s² · t² + 5.9 m/s · t · sin 40° + 2.4 m
Solving the quadratic equation:
t = 1.2 s
Now, we can calculate the x-component of the vector "r final" that is the horizontal distance traveled by the bicycle:
x = x0 + v0 · t · cos α
x = 0 m + 5.9 m/s · 1.2 s · cos 40°
x = 5.4 m
The BMX lands 5.4 m from the end of the ramp.
Have a nice day!
