Ok, here\'s more physics problems that needs explaination. Please help. Mathemat
ID: 586679 • Letter: O
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Ok, here's more physics problems that needs explaination. Please help.
Mathematical Equation entry to write the formula for the following quantities:
1 The components of the vector when you know magnitude and direction.
2 Position of an object (x) that moves with constant velocity during a period of time (t).
3 Position of an object (y) that moves with constant acceleration during a period of time (t).
1. If there's a spring gun. The end of the gun is at 1 meter above the ground. This gun is shoot horizontally with no air resistance. The cannon ball hits the ground 1.32 meters ahead from it was shoot. What is the cannon ball's initial velocity?
What are the datas given by the problem?
Initial position in the horizontal direction xi = Final position in the horizontal direction xf =
Initial position in the vertical direction yi = Final position in the vertical direction yf =
Initial velocity in the horizontal direction Vxi = Final velocity in the horizontal direction Vxf =
Initial velocity in the vertical direction Vyi = Final velocity in the vertical direction Vyf =
Initial time ti = Final time tf =
Acceleration in the horizontal direction ax = Acceleration in the vertical direction ay =
From the vertical data, find the time of flight t =
From the horizontal data and your time of flight, find your velocity in the horizontal direction Vx=
Can you consider this be the shooting velocity of your gun, regardless of the angle? Yes No
Why?
2. Now, with the same gun as in the previous problem. Using the initial velocity found and the same initial position (0, 1) meters. Find the final horizontal position if the cannon ball was shoot at 30° up to the air.
What are the Initial conditions:
Initial position in the horizontal direction xi = Initial muzzle velocity =
Initial position in the vertical direction yi = Final position in the vertical direction yf =
Initial velocity in the horizontal direction Vxi = Initial velocity in the vertical direction Vyi =
Initial time ti =
Acceleration in the horizontal direction ax = Acceleration in the vertical direction ay =
From the vertical data, find the time of flight t =
From the horizontal data and the time of flight, find the final position in the horizontal direction xf =
Explanation / Answer
xi = 0
xf = 1.32 m
yi = 1 m
yf = 0
vxi = ?
vxf = vxi
vyi = 0
vyf = ?
ti = 0
tf = ?
ax = 0
ay = -g
along vertical
(yf-yi) = vyi*t + 0.5*ay*t^2
(0-1) = (0) - (0.5*9.8*t^2)
t = 0.452 s
++++++++++++
xf-xi = vox*t + 0.5*ax*t^2
1.32 - 0 = vx*0.452 + 0
vx = 2.92 m/s
+++++++++++
No
as angle changes the x component vx changes
++++++++
xi = 0
vi = vo = 2.92 m/s
yi = 1
yf = 0
vxi = vo*costheta = 2.92*cos30
vyi = vo*sintheta = 2.92*sin30
ti = 0
ax = 0
ay = -g
along vertical
yf-yi = voy*t + 0.5*ay*t^2
-1 = (2.92*sin30*t) - (0.5*9.8*t^2)
t = 0.62 s
along horizantal
xf-xi = vox*t + 0.5*ax*t^2
xf-0 = (2.92*cos30*0.62)
xf = 1.56 m
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