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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