Question

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 4.3s for the dart to land back at the barrel.What is the maximum horizontal range of your gun?

Answer 1

When you fire the dart straight up you find the max hang time for the dart -> You can use
\[V_f = V_i -gt\] for the y direction to find Vi

This Vi will be the total Velocity allowed for the dart to be fired -> you than can shoot the dart at 45 degrees to maximize the range and apply this equation
\[d = \frac{v^2}{g}\] when v is the max velocity of the dart

Answer 2

ok so i got 42.14 m/s for Vi
im not sure what to do next, are you saying to plug in 42.12cos(45) for the v in the second equation?

Answer 3

To find the velocity, remember that when using
\[v_f=v_i-gt\]
Vf is zero when *half of the time it takes for the dart to land back in the barrel passes, so
\[v_i=(9.81m/s^2)(2.15s)\]
For the range part of the question, like @PhysicsGuru said, 45º will be ideal angle.
\[v_{xi}=v_i\cos 45º\]
\[v_{iy}=v_i \sin 45º\]
\[\cos 45º = \sin 45º\]
For the total time that the dart is in the air, its downward velocity at the end of the trajectory is equal in magnitude and opposite in direction to its upward velocity at the beginning of its trajectory, so time is given by
\[v_{yf}=v_{yi}-gt\]
\[ -v_{yi}=v_{yi}-gt\]
\[t=\frac{2v_y}{g}\]
and its horizontal displacement s given by
\[d=v_{xi}t\]
\[d=v_{xi}\Big(\frac{2v_y}{g}\Big)\]

\[d=\frac{(v_i \cos45º)( 2 v_i \sin45º)}{g}\]
\[d=\frac{v_i^2}{g}\]
Which, again, is what @PhysicsGuru had. So yeah, You just use the total initial velocity of the dart gun ^_^