Physics Question (1 Viewer)

[Jonneeh]

A person standing at the top of a hemispherical rock of radius R = 11.00 m kicks a ball (initially at rest on the top of the rock) to give it horizontal velocity v_i

1)What must be its minimum initial speed if the ball is never to hit the rock after it is kicked?

2) With this initial speed, how far from the base of the rock does the ball hit the ground?

_{Im a bit confused}

What im doing is just finding the time of flight t= sqrt(2*11/9.8) which is roughly 1.5s then dividing the horizontal radius 11m by this time to give me the "minimum initial speed for the ball to never hit the rock".

but them the second Q says find how far the ball will travel from the base which using my answer is 0.

So clearly im doing something wrong. Any thoughts?

 

[cheezcake]

You're making th incorrect assumption that if it hits 0 height when x is 9 meters then it would never have hit the hemisphere, but seeing as it travels in a parabolic path in fact it would have gone through the hemisphere, havn't done the question but I would try solve for [equation of projectile]>[x^2 + y^2 = 81] (equation of hemisphere - disregard negative)

 

[Jonneeh]

You're making th incorrect assumption that if it hits 0 height when x is 9 meters then it would never have hit the hemisphere, but seeing as it travels in a parabolic path in fact it would have gone through the hemisphere, havn't done the question but I would try solve for [equation of projectile]>[x^2 + y^2 = 81] (equation of hemisphere - disregard negative)

where did 9 come from?

 

[cheezcake]

where did 9 come from?

11 ssorry haha* my friend asked me the same question today and his was 9, change equation of circle accordingly

 

[Jonneeh]

11 ssorry haha* my friend asked me the same question today and his was 9, change equation of circle accordingly

but why would it follow a circular path?

 

[cheezcake]

Okay clearly tried to explain it too quickly.

So, what you did wrong was assuming that the value v(i) was always above the rock even though you only confirmed it at the end of it's flight time. I.E it held true for t= sqrt(22/g) but you didn't confirm it true for all values t

To do that I got the equations for both x and y displacement of the projectile ( a parabola obviously). Which was y = 11-(gt^2)/2 and x = v(i)t, I subbed the t value from the x equation into y to obtain an normal parabola equation (i.e convert parametric to cartesian form)

got y = 11 - (gx^2)/2v(i)^2

Now the equation of the rock (circle) we know from high school is x^2 + y^2 = 121

Now we need to solve these two equations so that the y value of the projectile Is ALWAYS greater than the y value of the rock (i.e the ball never hits the rock)

If you do that you get an inequlity relating x to v(i)

i got x^2 > (44v(i)^2)/g - (4v(i)^4)/g^2

So we know the equation holds true for ALL values of x, thus we can easily see that the only way to solve it is when x= 0

let x = 0 and simplify and you get v(i) = ~10.383 m/s

 

[cheezcake]

Okay clearly tried to explain it too quickly.

So, what you did wrong was assuming that the value v(i) was always above the rock even though you only confirmed it at the end of it's flight time. I.E it held true for t= sqrt(22/g) but you didn't confirm it true for all values t

To do that I got the equations for both x and y displacement of the projectile ( a parabola obviously). Which was y = 11-(gt^2)/2 and x = v(i)t, I subbed the t value from the x equation into y to obtain an normal parabola equation (i.e convert parametric to cartesian form)

got y = 11 - (gx^2)/2v(i)^2

Now the equation of the rock (circle) we know from high school is x^2 + y^2 = 121

Now we need to solve these two equations so that the y value of the projectile Is ALWAYS greater than the y value of the rock (i.e the ball never hits the rock)

If you do that you get an inequlity relating x to v(i)

i got x^2 > (44v(i)^2)/g - (4v(i)^4)/g^2

So we know the equation holds true for ALL values of x, thus we can easily see that the only way to solve it is when x= 0 (also with a little common sense you realise that if you take x =1,2,3 etc you get a different [v(i)> than]
answer all of which are larger than at 0, which is hence the minimum v(i))
let x = 0 and simplify and you get v(i) = ~10.383 m/s

Edit: there is probably much simpler, more elegant solution possible but i ceebs finding it out right now

 

[IamBread]

Well this is what I did.


That worked lol.