## Time, distance & displacement

1. Someone throws a stone horizontally from the top of a cliff and out to the sea. The top of the cliff is 45 metres above the sea. The initial speed of the stone is 30 m/s.

a) if the stone had been dropped vertically, how long would it have taken to reach the sea?

b) when the stone is thrown horizontally, how long does it take to reach the sea?

c) what horizontal distance does the stone travel before hitting the sea?

d) what is the total displacement of the store from it's original point?

2. I will be combining both equations and attempted solutions.

For a) : speed=distance/time
Therefore: s=d/t
st=d
t=d/s
t=45m/30ms
t= 1.5 s?

I am clueless about b, c and d

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
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 Mentor Your first answer is incorrect, since you did not include the acceleration due to the earth's gravity. Also, part (a) says the stone is dropped off the cliff. This implies that the initial velocity is zero. Have you come across the kinematic equations for constant acceleration? See this website.
 So: s=ut + 1/2 at^2 45 = 0*t + (1/2*10) *t^2 45= 0 + 5t^2 45=5t^2 45/5=t^2 9=t^2 square root of 9 = t therefore: t=3 secs?

Mentor

## Time, distance & displacement

Correct.

Now, for the other parts, you need to split up the motion into horizontal and vertical components. So, for (b) you want the time taken for the stone to reach s=0.
 I don't understand what you mean by 'time taken for the stone to reach s=0.'
 Mentor Well, if you take the sea as the origin of your coordinate system, and consider the vertical motion. Then, the time taken for the stone to hit the sea is the time taken for the stone to reach postion y=0.