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## Homework Statement

From the edge of a cliff, two stones are thrown at the same time, one vertically upwards and the other vertically downwards with the same velocity of 20 m/s. The second stone reaches the ground in 5 s. How long will the first stone be in air? Also find the height of the cliff.

## Homework Equations

## s=ut+\frac{1}{2}at^2 ##

## v=u+at ##

## The Attempt at a Solution

I got the height correctly.

**Assuming edge of the cliff as the origin,**

height h=-20*5-0.5*(-9.8)*(5^2)=-222.5 m.

For the first ball I found two displacements, one while going upwards (s

_{1}) and the other downwards (s

_{2}).

s

_{1}=20*t

_{1}-0.5*9.8*(t

_{1}

^{2})

s

_{2}=0*t

_{1}-0.5*9.8*(t

_{2}

^{2})=-4.9*(t

_{2}

^{2})

since s

_{1}=s

_{2}

∴ 20t

_{1}-4.9t

_{1}

^{2}=-4.9t

_{2}

^{2}.......(1)

also from v=u+at

0=20-9.8t

_{1}, thus t

_{1}=2.04 s

putting in (1) i got t

_{2}=2.04 s.

The total time the ball was in the air = t

_{1}+t

_{2}=4.08 s.

The book gives the answer as 9.08 s. :( Is it wrongly printed? The solution given is:

222.65=20*t+0.5*9.8*t

^{2}. Solving for t, t=9.08 s. But how can the displacement for the first ball be 222.65 m? That's the height of the cliff!!

Also, is it ok to write displacement s as,

s(u, t, a) = ut + 0.5*a*t

^{2}.

Will the examiner deduct marks for writing it like this? And u and a are constants so is it just s(t) or s(u, t, a) ?

Please help. Thanks.

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