Based on the gravity of the moon (~1.6 m/s^2) and the height of the canon (~5 m)...

[lilpeach]

Ok, so apparently I'm on the moon and I take six pictures of some canon shooting through the sky:

Frame 1
Altitude: 5.0
X-Position: 0.0
Y-Position: 0.0

Frame 2
A: 204.2
X: 60.3645
Y: 100.655

Frame 3
A:401.8
X:120.729
Y:201.31

Frame 4
A: 597.8
X: 181.0935
Y: 301.965

Frame 5
A: 792.2
X: 241.458
Y: 402.62

Frame 6
A: 985.0
X: 301.8225
Y: 503.275

a) Based on the gravity of the moon (~1.6 m/s^2) and the height of the canon (~5 m), you know that t seconds after being fired, the projectile should be at an altitude of approximately A(t) = -0.8t^2 + vt + 5 meters. Find v. (A(1)=204.2)

b) You know from common sense that the projectile's horizontal position, (x,y), should follow a line (i.e. the projectile's "shadow" will, of course, move in a straight line). Find the linear equation describing the horizontal motion of the projectile, x(t)=at+b and y(t) = ct+d.

c) How many seconds after being fired will the projectile land?

d) Where will the projectile land?

DUE TMRW NEED HELP pls

 

[stapel]

Ok, so apparently I'm on the moon and I take six pictures of some canon shooting through the sky:

Is it maybe a "cannon" (a machine which launches a projectile) rather than a "Canon" (a brand of camera)? Is it maybe the object shot off by the cannon which is "shooting through the sky"?

Frame 1
Altitude: 5.0
X-Position: 0.0
Y-Position: 0.0

Frame 2
A: 204.2
X: 60.3645
Y: 100.655

Frame 3
A:401.8
X:120.729
Y:201.31

Frame 4
A: 597.8
X: 181.0935
Y: 301.965

Frame 5
A: 792.2
X: 241.458
Y: 402.62

Frame 6
A: 985.0
X: 301.8225
Y: 503.275

What is the speed of the camera (Canon) taking pictures of the object launched (by the cannon)? What is the frame-rate? This information is necessary in order to determine "time" from "frame".

a) Based on the gravity of the moon (~1.6 m/s^2) and the height of the canon (~5 m), you know that t seconds after being fired, the projectile should be at an altitude of approximately A(t) = -0.8t^2 + vt + 5 meters. Find v. (A(1)=204.2)

From "A(1) = 204.2", I will guess that the camera is taking pictures at the extremely slow rate of one per second, so "Frame n+1" corresponds to "time t = n seconds".

Since you've been given this information, you can plug the given values into the given formula, and solve for the specified variable. Where are you stuck in this process?

b) You know from common sense that the projectile's horizontal position, (x,y), should follow a line (i.e. the projectile's "shadow" will, of course, move in a straight line). Find the linear equation describing the horizontal motion of the projectile, x(t)=at+b and y(t) = ct+d.

Are you expected to relate the x and y from this part of the exercise to the (t, (A(t)) in the previous part? If so, what methods have they given you for this? How far have you gotten in applying that information?

c) How many seconds after being fired will the projectile land?

Hint: What is the height, above ground level, when the things hits the ground?

d) Where will the projectile land?

At the time found in part (c), what is the horizontal displacement?

When you reply, please include a clear listing of your thoughts and efforts so far. Thank you!