Grey Council
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- Oct 14, 2003
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- 2004
here are a few questions that I'm stuck on:
A partical of mass m is projected vertically upwards from a point high above the ground. The particle experiences a resistance of magnitude mkv<sup>2</sup>. During its downward motion, the terminal velocity of the particle is V. Its initial velocity of projection is half this terminal velocity.
during upward motion:
V<sup>2</sup>a = -g(V^2 + v^2)
x = (V^2/2g)ln((5V^2)/(4(V^2 + v^2))
during downward motion:
V<sup>2</sup>a = g(V^2 + v^2)
Find the position of the particle relative to its projection point when it attains 60% of its terminal velocity.
I got an answer, but it looks heaps dodgy to me. I got:
(25 - 9V<sup>2</sup>)/ 20
hrm
second question (this one is a killer )
A particle of unit mass is projected vertically upwards against a constant gravitational force g and a resistance v/c, where v is the velocity of the particle and c is a constant. s is the distance travelled in time t; at t = 0, s=0, and v = c(h-g) where h is a constant. Write down the equation of motion of the particle. (ooo, btw, what exactly does it mean when it says equation of particle? Just what acceleration is?)
Find the time taken by the particle to reach its highest point, and find the height of that point.
The particle falls to its original position under gravity and under the same law of resistance. Will the time of descent be reater or les than the time of ascent? Give reasons for your answer.
and third question:
A particle of mass m kilograms is dropped from rest in a medium blah blah blah. resistance = 1/10 . mv^2
particle hits ground at ln(1 + rt2) seconds after dropping.
g = 10
Express v as a function of t. Hence find the speed with which the particle hits the ground. Give your answer in simplest exact form.
Find, in simplest exact form, the distance fallen by the particle before it hits the ground.
lol, out of approx 20 trial paper questions, these were the only resisted motion questions I couldn't do, so I guess i'm getting the hang of this. Blah to harder 3u motion though.
A partical of mass m is projected vertically upwards from a point high above the ground. The particle experiences a resistance of magnitude mkv<sup>2</sup>. During its downward motion, the terminal velocity of the particle is V. Its initial velocity of projection is half this terminal velocity.
during upward motion:
V<sup>2</sup>a = -g(V^2 + v^2)
x = (V^2/2g)ln((5V^2)/(4(V^2 + v^2))
during downward motion:
V<sup>2</sup>a = g(V^2 + v^2)
Find the position of the particle relative to its projection point when it attains 60% of its terminal velocity.
I got an answer, but it looks heaps dodgy to me. I got:
(25 - 9V<sup>2</sup>)/ 20
hrm
second question (this one is a killer )
A particle of unit mass is projected vertically upwards against a constant gravitational force g and a resistance v/c, where v is the velocity of the particle and c is a constant. s is the distance travelled in time t; at t = 0, s=0, and v = c(h-g) where h is a constant. Write down the equation of motion of the particle. (ooo, btw, what exactly does it mean when it says equation of particle? Just what acceleration is?)
Find the time taken by the particle to reach its highest point, and find the height of that point.
The particle falls to its original position under gravity and under the same law of resistance. Will the time of descent be reater or les than the time of ascent? Give reasons for your answer.
and third question:
A particle of mass m kilograms is dropped from rest in a medium blah blah blah. resistance = 1/10 . mv^2
particle hits ground at ln(1 + rt2) seconds after dropping.
g = 10
Express v as a function of t. Hence find the speed with which the particle hits the ground. Give your answer in simplest exact form.
Find, in simplest exact form, the distance fallen by the particle before it hits the ground.
lol, out of approx 20 trial paper questions, these were the only resisted motion questions I couldn't do, so I guess i'm getting the hang of this. Blah to harder 3u motion though.