Question

In a children’s story, a princess trapped in a castle wraps a message around a rock and throws it from the top of the castle. Right next to the castle is a moat. The initial velocity of the rock is 12 m/s[42° above the horizontal]. The rock lands on the other side of the moat, at a level 9.5m below the initial level. Air resistance is negligible. Calculate the rock’s time of flight? Calculate the width of the moat. Determine the rock’s velocity on impact with the ground.

Answer 1

Ok, the way to tackle this problem is:

Answer 2

I am sorry but I my computer just shot down in the middle of my answer and I lost it all. I won't write the whole thing again but here is a summed up version.

Answer 3

To find the time of flight you must add the time of flight on the parabolic motion (from point A to C) and the time of the rest of the motion (from C to D). The time from A to C you can find it by finding the time from A to B and multiplying it by two. You know that your velocity (v subscript By) on the y direction is zero in B, you know that the initial velocity on the y direction (v subscript Ay) must be twelve times the sine of fourty-two, and you know that the acceleration in the y direction is minus the gravitational acceleration (-g) in earth which is nine point eight meters per second squared, so you can derive the time from A to B doing this:\[v _{f}=v _{o}+at \rightarrow v _{By}=v _{Ay}+(-g)t _{AB} \rightarrow t _{AB}=\frac{ v _{Ay} }{ g }=\frac{ 12\frac{ m }{ s }\sin(42) }{ 9.8\frac{ m }{ s ^{2} } }=?\](I am sorry you will have to do the calculations yourself since I don't have a calculator with me)

Now to find the time from A to C you must multiply the time from A to B by two since it is a Parabolic motion.

To find the time from C to D it you only have to know the height of the throw (h) which is a given of 9.5m and the velocity at C in the y direction which is the same velocity on A in the y direction but with a negative sign.\[x _{f}=x _{o}+v _{o}t+\frac{ 1 }{ 2 }at ^{2}\rightarrow0=h+v _{Cy}t _{CD}+\frac{ 1 }{ 2 }(-g)t _{CD}^{2}\]\[t _{CD}=\frac{ -v _{Cy}\pm \sqrt{v _{Cy}^{2}-4(\frac{ 1 }{ 2 }(-g))(h)} }{ 2(\frac{ 1 }{ 2 }(-g)) }\]All you have left to do is to add the velocity from A to C and from C to D.\[\frac{ 2v _{o}sen \theta }{ g }+\frac{ v _{o}\pm \sqrt{v _{o}^{2}+4(\frac{ g }{ 2 })(h)} }{ -g }\](where h=9.5m, v subscript o=12 m per s, g=9.8 m per s squared, theta=42 degrees)

To find the distance of the moat you only have to multiply the velocity on A on the x direction times the total time since there is no acceleration on the x direction. the equation is:\[x=v _{Ax}t _{AD}=v _{o}\cos \theta(\frac{ 2v _{o}sen \theta }{ g }+\frac{ v _{o}\pm \sqrt{v _{o}^{2}+4(\frac{ g }{ 2 })(h)} }{ -g })\](where h=9.5m, v subscript o=12 m per s, g=9.8 m per s squared, theta=42 degrees)

You will find that the term in parentheses is the answer to the previous question (the total time) and the term outside is the velocity of the rock in A on the x direction.

Finally to find the rock's velocity at D (when it hits the ground) you must find the velocity on D on the y direction which is the velocity on the y direction at C (which is the velocity on the y direction at A with a minus sign) plus the aceleration time the time from C to D. Then you can find the rock's velocity at D by pythagoras between the velocity at D in x (which is the same in all points of the motion) and the velocity at D in y. You get:\[v _{D}=\sqrt{(v _{o} \cos \theta)^{2}+((-v _{o}\sin \theta)+(-g(\frac{ 2v _{o}sen \theta }{ g }+\frac{ v _{o}\pm \sqrt{v _{o}^{2}+4(\frac{ g }{ 2 })(h)} }{ -g })))^{2}}\](where h=9.5m, v subscript o=12 m per s, g=9.8 m per s squared, theta=42 degrees)

Any questions just ask :)