A ball is thrown w/initial speed vi at an angle 𝜃i with the horizontal

[Rudina]

Homework Statement

Find its initial speed

Relevant Equations

For d), I used V=square root Vy2+Vx2, but I got it wrong.

A ball is thrown with an initial speed vi at an angle 𝜃i with the horizontal. The horizontal range of the ball is R, and the ball reaches a maximum height R/8. In terms of R and g, find the following.
(a) the time interval during which the ball is in motion: Sroot(R/g) Correct
(b) the ball's speed at the peak of its path: (Sroot 4gR)/2 Correct
(c)the initial vertical component of its velocity Sroot(gR/4) Correct
(d) its initial speed
(e) the angle 𝜃i expressed in terms of arctan of a fraction.
(f) Suppose the ball is thrown at the same initial speed found in (d) but at the angle appropriate for reaching the greatest height that it can. Find this height.
(g) Suppose the ball is thrown at the same initial speed but at the angle for greatest possible range. Find this maximum horizontal range.

Thank you

 

[kuruman]

We cannot help you if you do not tell us exactly what answer you gave. If it is just V=square root Vy2+Vx2, then it is wrong because you are supposed to give the answer in terms of R and g.

Also, please take the time to learn how to use LaTeX for writing equations. It is a useful skill to have. Click the link LaTeX Guide, lower left, to see how it is done.

 

[Orodruin]

We cannot help you if you do not tell us exactly what answer you gave.

… and preferably the reasoning behind it!

 

[Rudina]

I apologize for that.
a) \sqrt {\frac R g} Correct!
b) {\sqrt {4gR}} \frac 2 Correct!
c) \sqrt {gR} \frac 4 Correct
d) The formula I used for the initial speed is this \sqrt {Vy^2+Vx^2}, and after I plugged in the information I got: (\sqrt {\frac {gR} 4}) +( \sqrt {\frac {4gR} {4}) = \sqrt {\frac {4gR} {4}. and I got it wrong.
I hope I wrote this right. I apologize. This is my first time in a forum, and I am still learning to navigate.

 

[Orodruin]

I apologize for that.
a) \sqrt {\frac R g} Correct!
b) {\sqrt {4gR}} \frac 2 Correct!
c) \sqrt {gR} \frac 4 Correct
d) The formula I used for the initial speed is this \sqrt {Vy^2+Vx^2}, and after I plugged in the information I got: \sqrt {\frac {gR} 4} + \sqrt {\frac {4gR} {4} = \sqrt {\frac {4gR} {4}. and I got it wrong.
I hope I wrote this right. I apologize. This is my first time in a forum, and I am still learning to navigate.

You need to put double hashes (##) around your equations for them to render properly.

Your expression on (d) renders as
$$
\sqrt {\frac {gR} 4} + \sqrt {\frac {4gR} {4}} = \sqrt {\frac {4gR} {4}}
$$
which surely cannot be what you intended?

Edit: I added a few missing }

 

[Rudina]

Yeah, I wrote that wrong. Anyway. It is almost 12:00 am here and I don't have much time to finish this question. Thank you for your patience.