1. ## Kinematics

Question:

A stone is thrown vertically downwards from the top of a cliff. Its height h metres above the beach below at time t seconds is given by.
h= (10- sqrt 5t) (10 + sqrt 5t)

show that acceleration is constant and state its value.

hence find the time when it hits the beach.

The answer is -10 ms-2 however I don't know how to manipulate the above expression.

sqrt 5t should be equal which would give a value of -25 and not 10, which brings me to wonder if the question is supposed to say sqrt 5 times t, would it make a difference? 5t^0.5 however when one multiplies the two and differentiates again would give -25.

2. Well, $(10-\sqrt{5t})(10+\sqrt{5t}) = 100-5t$. So the stone hits the ground at $t=20$.

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However, the problem probably means $h=(10-t\sqrt{5})(10+t\sqrt{5}) = 100-5t^2$, so the stone hits the ground at $t=\sqrt{20}=2\sqrt{5}$.

Thus, $velocity = \frac{dh}{dt}=-10t~m/s$ and $acceleration = \frac{d^2h}{dt^2} = -10~m/s^2$

3. Originally Posted by Mouseman
Question:

A stone is thrown vertically downwards from the top of a cliff. Its height h metres above the beach below at time t seconds is given by.
h= (10- sqrt 5t) (10 + sqrt 5t)

show that acceleration is constant and state its value.

hence find the time when it hits the beach.

The answer is -10 ms-2 however I don't know how to manipulate the above expression.

sqrt 5t should be equal which would give a value of -25 and not 10, which brings me to wonder if the question is supposed to say sqrt 5 times t, would it make a difference? 5t^0.5 however when one multiplies the two and differentiates again would give -25.
$h = (10 - \sqrt{5} t)(10 + \sqrt{5} t)$

$h = 0$ at $t = \frac{10}{\sqrt{5}}$ sec

expand the expression for h ...

$h = 100 - 5t^2$

$\frac{dh}{dt} = v = -10t$

$\frac{dv}{dt} = a = -10$