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Eine englische Aufgabe

Schwierigkeitsgrad: mittelschwere Aufgabe

Hinweis: Die folgende Aufgabe "A ball rolls off of a cliff" stammt von Professor Bloomfield aus England, dessen Originalseite leider nicht mehr im Netz steht; LEIFIphysik hat lediglich einige formale Änderungen vorgenommen.

Question: A ball rolls off of a cliff that is \(10\rm{m}\) high. If the ball lands \(5\rm{m}\) away from the edge,

a)How long does it take to hit the ground?

b)What was it's initial speed?

c)What will be it's final velocity (give both the magnitude and direction)?

Lösung einblendenLösung verstecken Lösung einblendenLösung verstecken

The ball does not fall in a straight line from A to B. It follows a parabolic path instead. Now, you may be nodding your head violently as if this were a too obvious statement, but you might also be feeling an urge to use that right triangle with a height of 10 and base of 5 to find some sort of angle. Whatever that angle is, it's certainly irrelevant to the problem at hand.

With any projectile motion problem, it's good practice to separate everything into \(x\) and \(y\) components. I usually do this with two columns. I also choose point A to be the point \((0|0)\).


\({x_0} = 0{\rm{m}}\)

\({x_{\rm{F}}} = 5{\rm{m}}\)

\({v_{0,x}} = ?\)


\({y_0} = 0{\rm{m}}\)

\({y_{\rm{F}}} = -10{\rm{m}}\)

\({v_{0,y}} = 0\frac{{\rm{m}}}{{\rm{s}}}\)

\({a_y} =-9.8\frac{{\rm{m}}}{{{{\rm{s}}^{\rm{2}}}}}\)

Note: I like to use \({a_y} =  - 9.8\frac{{\rm{m}}}{{{{\rm{s}}^{\rm{2}}}}}\) rather than using just \(9.8\frac{{\rm{m}}}{{{{\rm{s}}^{\rm{2}}}}}\) and having to remember to place my negatives in later. It's a really good practice to take care of all your negative values prior to plugging them into an equation. That way, you never have to mess with the equations.

Here's the equations that will see us through the remainder of the problem:

The motion of an object moving solely under the influence of gravity near the earth's surface is governed by \[\begin{array}{*{20}{c}}{x = {x_0} + {v_{0,x}} \cdot t}&{y = {y_0} + {v_{0,y}} \cdot t + \frac{1}{2} \cdot {a_y} \cdot {t^2}}\\{{v_x} = {v_{0,x}}}&{{v_y} = {v_{0,y}} + {a_y} \cdot t}\end{array}\] All you need are these equations, which are really the same for both \(x\) and \(y\) with the difference being that \({a_x} = 0\). All the worrying about negative signs should be done in the initial setup as we did above.

Now, plug in the values that were given. Since most of the known information is for the \(y\)-component, we'll start there. \[y = {y_0} + {v_{0,y}} \cdot t + \frac{1}{2} \cdot {a_y} \cdot {t^2} \Rightarrow  - 10{\rm{m}} = 0{\rm{m}} + 0\frac{{\rm{m}}}{{\rm{s}}} \cdot t + \frac{1}{2} \cdot \left( { - 9,8\frac{{\rm{m}}}{{{{\rm{s}}^{\rm{2}}}}}} \right) \cdot {t^2} \Leftrightarrow 2,04{{\rm{s}}^{\rm{2}}} = {t^2} \Rightarrow t = 1,43{\rm{s}}\] This answers the question about the time of flight. With this interesting tidbit, we can run over to the \(x\)-equation: \[x = {x_0} + {v_{0,x}} \cdot t \Rightarrow 5{\rm{m}} = 0{\rm{m}} + {v_{0,x}} \cdot 1,43{\rm{s}} \Leftrightarrow {v_{0,x}} = 3,5\frac{{\rm{m}}}{{\rm{s}}}\] and now we know what the initial horizontal speed was!


The final part asked about the final velocity of the ball. To do this, we'll refer to the equations for the individual component velocities: \[{v_x} = {v_{0,x}} = 3,5\frac{{\rm{m}}}{{\rm{s}}}\] \[{v_y} = {v_{0,y}} + {a_y} \cdot t = 0\frac{{\rm{m}}}{{\rm{s}}} + \left( { - 9,8\frac{{\rm{m}}}{{{{\rm{s}}^{\rm{2}}}}}} \right) \cdot 1,43{\rm{s}} =  - 14\frac{{\rm{m}}}{{\rm{s}}}\] Since these are the components of the final velocity, we'll need to slap a quick Pythagorean theorem on them to get the magnitude of the total velocity: \[v = \sqrt {{v_x}^2 + {v_y}^2}  = \sqrt {{{\left( {3,5\frac{{\rm{m}}}{{\rm{s}}}} \right)}^2} + {{\left( { - 14\frac{{\rm{m}}}{{\rm{s}}}} \right)}^2}}  = 14,4\frac{{\rm{m}}}{{\rm{s}}}\]Finally, to get the direction, we'll apply a little trig to find the value of \(\Theta \) (Übersetzungshilfen: "trig" bedeutet Trigonometrie, "adj" bedeutet Ankathete und "hyp" bedeutet Hypotenuse) \[\cos \left( \Theta  \right) = \frac{{{\rm{adj}}}}{{{\rm{hyp}}}} = \frac{{{v_x}}}{v} = \frac{{3,5\frac{{\rm{m}}}{{\rm{s}}}}}{{4,4\frac{{\rm{m}}}{{\rm{s}}}}} = 0,24 \Rightarrow \Theta  = \arccos \left( {0,24} \right) = 76^\circ \]

Grundwissen zu dieser Aufgabe


Waagerechter und schräger Wurf