a heavy sphere is thrown 45 degrees upwards. it spends twice as long descending as ascending, because it's been thrown off a 30-metre-high cliff. it then hits the ground, and with each bounce loses 80% of its energy. will the ball hit someone standing 70metres from the cliff?
assume g = 10 m/s/s

this is all the question says. i think i've worked out the initial speed, but i'm not sure
i am also assuming no air resistance, each time the ball hits the ground the collision is elastic, and that the person standing 70m away is infinitely tall, so that the ball cannot bounce over

thanks

2. Originally Posted by cassius
a heavy sphere is thrown 45 degrees upwards. it spends twice as long descending as ascending, because it's been thrown off a 30-metre-high cliff. it then hits the ground, and with each bounce loses 80% of its energy. will the ball hit someone standing 70metres from the cliff?
assume g = 10 m/s/s

this is all the question says. i think i've worked out the initial speed, but i'm not sure
i am also assuming no air resistance, each time the ball hits the ground the collision is elastic, and that the person standing 70m away is infinitely tall, so that the ball cannot bounce over

thanks
As far as I can see in order to do this will require that you make some assumptions about what is happening at each bounce, but these do not look reasonable.

CB

3. this is all the question says. i think i've worked out the initial speed, but i'm not sure
There are many ways to find initial velocity....I would like to see your way ...in case of an error ..it will be pointed out.. so its my request that you kindly post your attempt at initial velocity and get your concept cleared..
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Here is a hint at the main punch of your question.

After every collision the velocity will become $(\frac{1}{\sqrt(5)})$th of its value just before collision

Range of a projectile with velocity v angle $\theta$ is given by

$R= \frac{v^2 sin(2\theta)}{g}$

When you will add the range it will be of the form

$R + R_1 +R_2+R_3+.......= R + \frac{{v_1}^2 sin(2\theta)}{g}+\frac{{v_2}^2 sin(2\theta)}{g}+\frac{{v_3}^2 sin(2\theta)}{g}+......$

$\theta$ will remain same and would be greater than 45 degree in this case

$R + R_1 +R_2+R_3+.......= R + \frac{{v_1}^2 sin(2\theta)}{g}+\frac{{v_1}^2 sin(2\theta)}{5g}+\frac{{v_1}^2 sin(2\theta)}{5*5g}+......$

This will be an infinite GP the total sum of which is given by a/(1-r)

$R_{complete}= R + R_1 +R_2+R_3+.......= R + \frac{{v_1}^2 sin(2\theta)}{g(1-\frac{1}{5})}$
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You wont need to consider a 1000 feet giant for this question

Take Care