model rocket flies horizontally off the edge at a velocity of 20m/s. if canyon below is 90m deep, how far from the edge does the rocket land?
it's a physics question and it is about vectors??
- WhomeLv 74 weeks ago
Let's just assume that the rocket is no longer propelled by exhaust when it departs the cliff edge and that air resistance can be neglected.
Use the general equation for position in time with constant acceleration
s = s₀ + v₀t + ½at²
If up is the positive direction and the top of the cliff is origin.
in the vertical direction initial position and vertical velocity can be considered zero.
-90 = 0 + 0t + ½(-9.8)t²
t = √(2(-90)/(-9.8))
t = 4.286 s to hit the canyon floor.
in the horizontal, our equation looks thus
s = 0 + 20(4.286) + ½(0)t²
s = 85.714...
s = 86 m from the point below the cliff rim.
I hope this helps.
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- SkyLv 74 weeks ago
There's not enough information. Once clear of the edge or any surface supporting it, if there is any imbalance in the rocket's weight distribution or thrust line compared to center of gravity, the rocket can begin to tilt either up or down, drastically changing its final trajectory. If it has a guide stick like so many fireworks rockets, if the stick is on top the thrust line will be below that mass, causing the rocket to rotate upwards, and if the stick is on the bottom the thrust line will be above it, causing it to rotate downwards. It might even start to tilt upwards, and then as the fuel burns and the CG shifts forward, it begins to tilt downwards. If it's the kind with guide fins, they would have to be very precisely mounted in order for there to be no sideways lift that would cause it to tilt up or down (or left or right) and instead have an absolutely perfect aerodynamic line compared to the rocket and its thrust vector. Also, a rocket taking off will keep accelerating until thrust = drag, making a speed of 20 m/s only a single instantaneous velocity. This problem is impossible to predict with so little information provided.