Example
A boy kicks a ball horizontally over the edge of a cliff with a speed of \(6ms^{-1}\) as shown in the diagram.
The ball hits the surface of the water \(3s\) later.
(a) Calculate the vertical speed of the ball when it reaches the water's surface.
(b) How high is the cliff?
(c) How far from the foot of the cliff will the ball land?
Answer
Horizontal motion
\(v=6\,ms^{-1}\)
\(t=3s\)
Vertical motion
\(u=0\,ms^{-1}\) (the ball does not move vertically until it goes over the edge)
\(a=9.8\,ms^{-2}\) (this is the downwards acceleration due to gravity)
\(t = 3 s\)
\(v = ?\)
a) Use only the vertical information.
\(v = u + at\)
\(v = 0 + (9.8 \times 3)\)
\(v=29.4\,ms^{-1}\)
b) The height of the cliff is simply the vertical distance travelled by the ball. If we draw a speed-time graph using the vertical motion, we can calculate the vertical distance travelled.
The ball has an initial vertical velocity of \(0 ms^{-1}\) and accelerates uniformly over \(3s\) to reach a final vertical speed of \(30ms^{-1}\).
The area under the graph is the vertical distance travelled:
\(Area = 0.5 \times base \times height\)
\(= 0.5 \times 3 \times 29.4\)
\(= 44.1 m\)
So the height of the cliff is \(44.1\,m\).
c) The ball travels through the air for \(3s\) before it reaches the sea. The horizontal speed of the ball is constant, so we can use:
\(d = vt\)
\(d = 6 \times 3\)
\(d = 18 m\)
The ball is \(18\,m\)from the base of the cliff when it lands in the sea.
