Question

Physics Help! If George Washington threw a silver dollar across a river which is 300m wide, with what speed did he have to throw the coin? Assume a 45 degree angle. I got 27.12 m/s correct me if I'm wrong. This problem was difficult

Answer 1

Does anybody know it?

Answer 2

Please tell me you know it stormfire

Answer 3

In order to answer this we need to remember some physics. The f1rst thing to remember is that we are dealing with projectile motion, which means that for an object thrown at an angle with a given velocity, that velocity will have a vertical and a horizontal component.

The second thing we will recall is that only the vertical component is effected by gravity, and that the horizontal component will be remain constant.

The third thing we shall recall is that for projectile motion across a flat plane (that is that the launch and landing points are at the same height), that the arc is symmetrical about the point of maximum height. What this means is that the velocity of the vertical component at launch will be identical to the vertical component of velocity at landing. Consequently, because the horizontal component is constant, then total velocity upon landing is identical to the launch velocity.

Fourthly, because the launch angle is 45 degrees, the magnitude of the vertical component of velocity at launch will be identical to the magnitude of the horizontal component at launch, and consequently the same when it lands (i.e. it will land at an angle of 45 degrees).

Fifthly, the range of a particle is governed solely by its time in the air (time of flight), and its horizontal velocity.

Lastly, the time of flight for it to cover the complete range (i.e. from it being launched to it landing) will be twice the time it takes for the particle to reach its maximum height (which will be at a point equating to half the range).

So by recalling each of these, we can now solve the problem. To simplify things, we are going to invoke the symmetry of the arc, and only consider the motion from the second half of the arc (from the time after the coin has reached its highest point to its landing).

We know that the range \(R\) of the particle is equal to its horizontal velocity \(V_h\) times its time of flight (which we also recall is twice the time \(t\) it takes for the particle to reach its highest point from launch, or for it to fall to the ground from its highest point). We can therefore express that the half range \(R_{1/2}\) will be the product of the fall time from its highest point and the horizontal component or \[R_{1/2}=V_ht\]Now a particle at its highest point and falling under the acceleration of gravity, will have an initial vertical velocity of \(U_v=0\), and a vertical velocity of \(v_v\) at impact, which can be calculated from the equation of motion \[V_v=U_v+at\]Since \(U_v=0\) then the above simply becomes \[V_v=at\]

If we now combine the range equation with the equation of motion to eliminate \(t\) to get \[V_v=a\frac{R_{1/2}}{V_h}\] or \[V_hV_v=aR_{1/2}\]

Now recall that because the launch angle is at 45 degrees, and hence the landing angle is the same, then the magnitude of the horizontal and vertical velocities at these points are equal i.e. \(V_v=V_h\) and the above becomes \[(V_h)^2=aR_{1/2}\] and hence \[V_h=\sqrt{aR_{1/2}}\]

So if the range is 300 m, then the half range \(R_{1/2}=150\) m, and taking \(a=9.81\) m/s\(^2\) we get that \(V_h=38.36\) m/s. But remember that this is the horizontal component, but it will also equal the vertical component. to find teh actual velocity of launch, we must recombine the two components, which can be done using Pythagoras, since the two velocities are at right angles to each other, with the combined velocity forming the hypotenuse of this right angled triangle. Therefore \[V=\sqrt{(V_h)^2+(V_v)^2}=\sqrt{2(V_h)^2}=\sqrt{2(38.36)^2}= 54.25\rm{m/s}\]

Judging by your answer, you appear to have worked through some of the physics correctly, but have misplaced a factor of 2 somewhere.