Ok so here is the situation. My teacher gave us a nerf gun and said that we have a week to find an equation to determine where it will land on testing day where she gives us a certain angle and we have to put the angle into the equation or whatever and find guess the spot the dart will land. Can anybody help?
http://www.physicsclassroom.com/class/vectors
this may website could help you.If you not learn vector component yet,you could start form lesson 1.
If you did,you can start from lesson 2.
And to determine where the nerf gun will land,you need the initial velocity(take off speed) of the nerf gun.The height from the gun to the ground that you strat will also affect where it land.
What you need to see are the kinematics equations, which you can read about here: http://www.physicsclassroom.com/Class/1DKin/u1l6a.cfm
The one that applies here states that distance equals initial velocity plus half acceleration times time squared, that's
d = i + ( a * t^2 ) / 2 when i equals initial velocity.
Assuming no air resistance, your acceleration on the x axis will be constant, and your acceleration on the y axis will be that of gravity, that's 9.8 m/s^2. The total velocity equals velocity on the x axis plus velocity on the y axis. We want to relate the total velocity, a vector, to the velocity on the x axis, another vector, with an angle (E for angle of elevation) between the two. Because we're working on the Cartesian plane which features x and y axes perpendicular to each other, we can use a trigonometric ratio, cosine E equals velocity on the x axis over total velocity. Use this to find the initial velocity on the x axis.
What you need to do now is to solve for time on the y axis and plug that in to find distance on the x axis. This is made simple given the fact that distance (d) will be zero on the y axis (what comes up must come down), so we substitute that into our equation and solve. 0 = i + ( -9.8 * t^2 ) / 2 Since initial velocity (i) is a constant, we can easily find that through experimentation.* Before we do either of those thing, our equation for distance on the x axis will look like this: d = i + ( 1 * t^2 ) / 2 because we're assuming no air resistance.
*For example by shooting the nerf gun, timing it, measuring the distance and working backward from there using your equation. You could even do this at varying angles to try to approximate air resistance.
I'm a bit rushed in writing this as I'm on a library computer and it's getting close to closing time, so I apologize for an discrepancies, please pm me if you find me in error. If you liked this answer please follow me on Quora: https://www.quora.com/Azeezah-M
I am sorry,guest.Even I just started to learn physics in this semester,but I believe you had done soemthing wrong.Yes, I agree that you can use \(X=Xo+Vxo*t+1/2*Ax*t^2\)determine the where it will land.
But you misstyped that equation,and this equation can only use to determine the horizontal displacement of the nerf gun.It should be \(d=i*t+1/2*a*t^2\) .You can only determine horizontal displacement when time is given.
How about verical displacemnt(height)?You can only use your equation when throw the gun in 0 degree betweeen or parallel to the ground.Because \(Vxo=V*cos(theta)\)
Theta is the angle between where you track and horizontal surface.
Vxo=V*cos(0)=V inthis case Vx=V
But do think the teacher always give us an theta equal 0.NO!It could be any angle,such as 60 degree.
Then the horizontal initial velocity is Vxo=V*cos(60)=V/2
Like this
and Vy=sin(theta）*V=sqrt(3)*V/2
again,theta can be angle.
A projectile is an object upon which the only froce acting is gravity.Gravity is gravitation field contrast unique maaasive object cause by the curature from wrap in space time.
The ration(Fnet/m） is sometimes called the gravitional field strength and is expressed as 9.8N/Kg(9.8m/s^2).That is if an object moving upward,its acceleration is -9.8m/s^2 up,it will decrease its velocity to 0 and then drop.
acceleration is change in velocity in a given amount of time
acceleration =(final velocity -initial velocity)/time a=(V-Vo)/t
rearrange we have V=Vo+at Apply it to vertical velocity
We have Vy=Vyo+Ay*t
Vy is the vertical final velocity. Vyo is the vertical initial velocity. Ay is the vertical accelreation
t=time
Since Vyo=sin(theta)*V so Vy=V*sin(theta)+Ay*t
and the equation for Y(vertical displacement) is
Y=Yo+Vyo*t+1/*Ay*t^2
You can find the equation for hroizontal displacemt-vertical displacemt(Y-X)( eqution of trajectory) base on the formula that I listed.And you can find more information from physicsclassroom.
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