Projectile Motion Kinematics: A catapult launches a rock....

[DracoMalfoy]

Homework Statement

A catapult launches a rock with an initial velocity of 50m/s at 35 degrees above the horizontal. The rock must clear 3.5m at 10m away in order to make it over a castle wall to get to the enemy. Do the knights succeed? By how much do they clear the wall or miss clearing the wall? (Ans: 3.1m)

Homework Equations

X-direction[/B]

• Vfx=Vix: Vicosθ
• a= 0m/s
• Δx=Vix(t)

Y-direction

• Viy=Visinθ
• Vfy-Visinθ+gt
• Δy=(visinθ)(t)+1/2gt^2
• Vfy^2=(Visinθ)^2+2gΔy
• Δy=1/2(Vfy+Visinθ)(t)
• a=-9.8m/s^2

The Attempt at a Solution

I drew a picture. TheViy is 28.6. The Vix and Vfx is 40.9. The acceleration in the x-direction is zero. In the y-direction its -9.8m/s^2. Am I supposed to the Δy and then subtract it from 3.5 which is the height of the castle at 10m away?

 

[tnich]

Am I supposed to the Δy and then subtract it from 3.5 which is the height of the castle at 10m away?

It seems like that would answer the questions.

 

[alejandromeira]

You have Y as function of (t)

Why don't calculates Y value at x=10m? (You must do a previous calculation with the X equation)

 

[Cutter Ketch]

• Viy=Visinθ

• Vfy-Visinθ+gt

You are welcome to choose up or down as positive, but you must be consistent. In the first equation you have chosen up as positive. In the second equation (I assume that “-“ was supposed to be an “=“ right?) you have gt as positive which would suggest down is positive. Fix one or the other.

 

[SammyS]

Homework Statement

A catapult launches a rock with an initial velocity of 50m/s at 35 degrees above the horizontal. The rock must clear 3.5m at 10m away in order to make it over a castle wall to get to the enemy. Do the knights succeed? By how much do they clear the wall or miss clearing the wall? (Ans: 3.1m)

Homework Equations

X-direction[/B]

• Vfx=Vix: Vicosθ
• a= 0m/s
• Δx=Vix(t)

Y-direction

• Viy=Visinθ
• Vfy-Visinθ+gt
• Δy=(visinθ)(t)+1/2gt^2
• Vfy^2=(Visinθ)^2+2gΔy
• Δy=1/2(Vfy+Visinθ)(t)
• a=-9.8m/s^2

The Attempt at a Solution

I drew a picture. The Viy is 28.6. The Vix and Vfx is 40.9. The acceleration in the x-direction is zero. In the y-direction its -9.8m/s^2. Am I supposed to the Δy and then subtract it from 3.5 which is the height of the castle at 10m away?

You asked:

" Am I supposed to the Δy and then subtract it from 3.5 which is the height of the castle at 10m away? "​

If you plan to do that, how do you get a value for Δy ? What would you use for a value for time, t ?

Regarding Post #4 by @Cutter Ketch : It appears that you are using a negative value for g, namely −9.8 m/s², so your sign conventions should work just fine.

 

[Taylor_1989]

If given these two equations

$y-y_0=v_{y_0}t-\frac{1}{2}gt^2$ [1]

where
$y$ is the final height above ground
$y_0$ is the initial height of the ball (ground level)
$v_{y_0}$ is the initial velocity of the ball (ground level)
$t$ is the time of flight of ball
$g$ is gravity from what I have read you assign as $-g$

and the other equation

$x=v_{x}t$ [2]
where
$x$ is the displacement of the ball at a given time
$v_x$ is the horizontal velocity

can you see away of substituting one of these equations into the other, what dose equation [1] and [2] have in common?