- #1
MichaelTam
- 93
- 6
- Homework Statement
- Test
- Relevant Equations
- ## a[t]= A - B t^2 ##
A person standing a distance š from the rocket launch site shoots a projectile at š”=0 at an initial speed š£0 at an angle š0 with respect to the horizontal as shown in the figure above. The projectile hits the rocket just when the rocket reaches its maximum height. The downward gravitational acceleration is š.
(Part b) Find an expression for the tangent of the angle tan(š0) in terms of š“, šµ, š, and š. Do not use š£0 in your answer.
,In part a, I use the formula for finding ##t_f = \sqrt {3 A/B}## where ##t_f## is the time the rocket approach to the maximum height where the velocity will become zero due to the gravity on the rocket, where now the question canāt let me use ##v_0## as one of the answerās variable ,according to the Pythagoreanās Therom , ##v_0^2=v_0,y^2+v_0,x^2##or ##\tan \theta=\frac {v_0,y} {v_0,x} ##, so my first goal is to find out the ##v_0## firs.
My assumption according to the question is the initial position of the rocket is ## x=d, y = 0 ##
the initial position of the stone is ## x=0 , y=0 ##
The velocity of the rocket approaches the highest point,## v t_f = A t_f - \frac {B t_f^3} {3} = 0##
However, I donāt have the information of the stone velocity when it approaches ##t_f##
But the height of the stone and the rocket should be the same at ##t_f## so...
Velocity of the stone at time equals to ##t_f##
## \frac {A t_f^2} {2} - \frac {B t_f^4} {4} = v_y,0 - g t_f^2/2##
after substitute the ##t_f ##equation and simplify ,I get ##v_y,0= (\sqrt {48 A g ({g-A}) + 3 A })/4##
For ##v_x,0##, ## v_x,0=d/t_f##
substitute ##t_f## equation ,I get
##v_x,0= \frac {d \sqrt {3AB}} {3 A}##
using ##\tan \theta=\frac {v_0,y} {v_0,x} ##, and substitute the result and simplify,I get
##\tan \theta = \frac {\sqrt {144 A^2 g^2 - 144 A^3 + 9 A^2 B}} {4Bd}##
I donāt sure that if my step and assumption is correct, because I am not confident on two dimensions, especially the projectile because I have few confident on dealing with the algebra too...
(Part b) Find an expression for the tangent of the angle tan(š0) in terms of š“, šµ, š, and š. Do not use š£0 in your answer.
,In part a, I use the formula for finding ##t_f = \sqrt {3 A/B}## where ##t_f## is the time the rocket approach to the maximum height where the velocity will become zero due to the gravity on the rocket, where now the question canāt let me use ##v_0## as one of the answerās variable ,according to the Pythagoreanās Therom , ##v_0^2=v_0,y^2+v_0,x^2##or ##\tan \theta=\frac {v_0,y} {v_0,x} ##, so my first goal is to find out the ##v_0## firs.
My assumption according to the question is the initial position of the rocket is ## x=d, y = 0 ##
the initial position of the stone is ## x=0 , y=0 ##
The velocity of the rocket approaches the highest point,## v t_f = A t_f - \frac {B t_f^3} {3} = 0##
However, I donāt have the information of the stone velocity when it approaches ##t_f##
But the height of the stone and the rocket should be the same at ##t_f## so...
Velocity of the stone at time equals to ##t_f##
## \frac {A t_f^2} {2} - \frac {B t_f^4} {4} = v_y,0 - g t_f^2/2##
after substitute the ##t_f ##equation and simplify ,I get ##v_y,0= (\sqrt {48 A g ({g-A}) + 3 A })/4##
For ##v_x,0##, ## v_x,0=d/t_f##
substitute ##t_f## equation ,I get
##v_x,0= \frac {d \sqrt {3AB}} {3 A}##
using ##\tan \theta=\frac {v_0,y} {v_0,x} ##, and substitute the result and simplify,I get
##\tan \theta = \frac {\sqrt {144 A^2 g^2 - 144 A^3 + 9 A^2 B}} {4Bd}##
I donāt sure that if my step and assumption is correct, because I am not confident on two dimensions, especially the projectile because I have few confident on dealing with the algebra too...