- #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.

,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...

**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...