Verified answer

I'm going to go through this in variable form and plug in the numbers at the end so that we get part b for free.

Assumptions:

* The ball was released from rest

* There's no friction or air resistance

* The changes in acceleration due to changing height are negligible (very fair for 5m)

(Also, define up as the positive direction for the vertical.)While the ball is falling, we say that it has a constant acceleration of g downwards. This means we can use our constant-acceleration kinematics equations to find the final velocity:v^2 = v0^2 + 2adv^2 = 0 + 2(-g)(-h) = 2gh

v = -sqrt(2gh)  (downwards)

Then, we have an impact with the ground and we get a new velocity of 4sqrt(2gh)/5 upwards. (just 80% of our other velocity going upwards)

At the maximum height, we should have zero velocity: if the ball was moving up then we wouldn't be at maximum height, and if it was moving down then it would already have begun to fall and must be some distance below the max height.

So, using our kinematics equations again we get:

v^2 = v0^2 + 2ad

0 = (4sqrt(2gh)/5)^2 + 2(-g)y

2gy = 16/25 * 2gh

y = 16/25 * h

So if h = 5 m, y = 16/25 * 5 m = 3.2 m

Also, notice that our value for g appears nowhere in our final formula, so our answer is not dependent on g.

Hope this helps!

When the ball is dropped we have a = g, s = 5, u = 0 and v = v

so, using v² = u² + 2as we get:

v² = 10g

so, velocity on hitting the ground is v = √(10g)

Now considering upward motion we have:

u = 0.8√(10g), a = -g, s = s and v = 0...at maximum height

so, again using v² = u² + 2as we get:

0 = 0.8²(10g) - 2gs...(1)

i.e. 2s = 6.4

Hence, s = 3.2 metres

Note: From (1) the gravitational constant g cancels, hence final value for s is independent of g

:)>

You posted this in the totally wrong forum.