Question

A diver running 2.5 m/s dives out horizontally from the edge of a vertical cliff and 3.0 s later reaches the water below. how high was the cliff and how far from its base did the diver hit the water?

Answer 1

Answer:

a) Height of cliff = 44.145 meter

b) 7.5 meter far from its base the diver hit the water.

Explanation:

 We have equation of motion , $s= ut+\frac{1}{2} at^2$, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Diver's vertical motion:

   Initial velocity = 0 m/s, acceleration = 9.8 $m/s^2$, we need to calculate displacement when time is 3 seconds.

  $s= ut+\frac{1}{2} at^2\\ \\ =0*3+\frac{1}{2} *9.81*3^2\\ \\ =44.145meter$

 So height of cliff = 44.145 meter.

b) Diver's horizontal motion:

Initial velocity = 2.5 m/s, acceleration = 0 $m/s^2$, we need to calculate displacement when time is 3 seconds.

$s= 2.5*3+\frac{1}{2}*0*3^2\\ \\ =7.5meter$

So 7.5 meter far from its base the diver hit the water.

Answer 2

Answer:

Height of cliff is  $44.1\;\rm{m}\;\&\;7.5\;m$ far from its base the diver hit the water.

Explanation:

Given: A diver running $2.5\;\rm{m/s$ dives out horizontally from the edge of a vertical cliff and $3.0\;\sec$ later reaches the water below.

As per metioned in question:

Intial velocity of diver $(u)$ is $0\;\rm{m/s$.

Acceleration due to gravity $a=g=9.8\;\rm{ms^{-2$.

Using laws of motion: $s=ut+\frac{1}{2}at^2$

$s=0\times3+0.5\times9.8\times 3^2\\s=0.5\times9.8\times9\\s=44.1\; \rm{m$

So,Height of cliff $=44.1\;\rm{m$.

Now, horizontal motion is calculated as:

Intial velocity of diver $(u)$ is $2.5\;\rm{m/s$.

Acceleration due to gravity $a=0\;\rm{ms^{-2$.

Using laws of motion: $s=ut+\frac{1}{2}at^2$

$s=2.5\times3+\frac{1}{2}\times0\times3^2\\s=7.5\;\rm{m$

Hence, Height of cliff is  $44.1\;\rm{m}\;\&\;7.5\;m$ far from its base the diver hit the water.

Learn more about Laws of motion here:

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