You buy a plastic dart gun, and being a clever physics student you decide to do a quick calculation to find its maximum horizontal range. You shoot the gun straight up, and it takes 4.0 s for the dart to land back at the barrel. What is the maximum horizontal range of your gun?
Please note that the answer is 40m and that the three basic physics kinematics equations should be used. Like 1/2at^2+initial velocity (time)+initial position=final position.
Thank you!
Please note that the answer is 40m and that the three basic physics kinematics equations should be used. Like 1/2at^2+initial velocity (time)+initial position=final position.
Thank you!
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Hello Shoppy,
The trick to this question is breaking the flight into two pieces. For the first part, we could consider only the half of the flight that goes up or only the half of the flight that comes down. The time for each part would be 2 seconds, half of the total time. If you choose the first half of the flight, Vi is unknown, Vf = 0 m/s and a = 9.81 m/s^2. If you choose the second half of the flight Vi = 0 m/s, and Vf = unknown (this will equal the speed that the dart left the gun originally, just in the opposite direction.).
Either way we choose the same motion equation, the one without distance,
Vf = Vi + a*t
Vi = Vf - a*t
For first half of flight:
Vf = 0 m/s
Vi = ??
t = 2 sec
a = -9.81 m/s^2
Vi = Vf - a*t
Vi = -a*t [eq 1]
Vi = - (-9.81 m/s^2) * 2 sec
Vi = 19.62 m/s
Now the maximum range of a projectile is achieved if it is fired at a 45 degree angle, I hope that you have covered this.
Now we know the velocity that the dart leaves the gun, 19.62 m/s, for the second half of the problem, lets call this V.
Since we are shooting at 45 degrees, the components are:
Vx = V cos 45
Vy = V sin 45
This is how we will do the second part of the problem. We can get the total time of flight by working with the y direction, then feed this time into the x direction to find the total distance traveled horizontally.But to get an equation with only one unknown, we have to take just half of the vertical flight. That is because we know the velocity in the y direction at the top of the arc is zero.
The trick to this question is breaking the flight into two pieces. For the first part, we could consider only the half of the flight that goes up or only the half of the flight that comes down. The time for each part would be 2 seconds, half of the total time. If you choose the first half of the flight, Vi is unknown, Vf = 0 m/s and a = 9.81 m/s^2. If you choose the second half of the flight Vi = 0 m/s, and Vf = unknown (this will equal the speed that the dart left the gun originally, just in the opposite direction.).
Either way we choose the same motion equation, the one without distance,
Vf = Vi + a*t
Vi = Vf - a*t
For first half of flight:
Vf = 0 m/s
Vi = ??
t = 2 sec
a = -9.81 m/s^2
Vi = Vf - a*t
Vi = -a*t [eq 1]
Vi = - (-9.81 m/s^2) * 2 sec
Vi = 19.62 m/s
Now the maximum range of a projectile is achieved if it is fired at a 45 degree angle, I hope that you have covered this.
Now we know the velocity that the dart leaves the gun, 19.62 m/s, for the second half of the problem, lets call this V.
Since we are shooting at 45 degrees, the components are:
Vx = V cos 45
Vy = V sin 45
This is how we will do the second part of the problem. We can get the total time of flight by working with the y direction, then feed this time into the x direction to find the total distance traveled horizontally.But to get an equation with only one unknown, we have to take just half of the vertical flight. That is because we know the velocity in the y direction at the top of the arc is zero.
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