calc 3- projectile motion

[jenquin12]

A golf ball is hit from ground level at an angle of 40° degrees above ground level. The terrain is level. The ball lands 194.55 meters away.What was the initial speed of the golf ball?
For that same initial speed, find two angles of departure for which the ball will travel to a distance of 194.55 meters. ( Answer in degrees)

No idea how to start this!! help!

 

[Deleted member 4993]

A golf ball is hit from ground level at an angle of 40° degrees above ground level. The terrain is level. The ball lands 194.55 meters away.What was the initial speed of the golf ball?
For that same initial speed, find two angles of departure for which the ball will travel to a distance of 194.55 meters. ( Answer in degrees)

No idea how to start this!! help!

Let the initial velocity = u → initial horizontal velocity = u*cos(Θ) ................ Now continue

 

[jenquin12]

okay so I know that vo= vocos(f) + vosin(f) and they give you initial conditions for f=40 degrees and xfinal, or where the ball lands, is 194.55 meters and I know that speed is the magnitude of velocity but I keep getting 1.5867 but that is not correct. Any ideas where I went wrong?

 

[Deleted member 4993]

okay so I know that vo= vocos(f) + vosin(f) and they give you initial conditions for f=40 degrees and xfinal, or where the ball lands, is 194.55 meters and I know that speed is the magnitude of velocity but I keep getting 1.5867 but that is not correct. Any ideas where I went wrong?

I cannot tell where you went wrong - because I cannot see your work.

 

[jenquin12]

I just did magnitude of v=vocos(f) + vosin(f) but there is also another equation that includes acceleration, which acceleration is just a = -g which would then be used as F=ma= m(dr^2/dt^2)=-mg. d^2r/dt^2=-g therefore dr/dt= -gt + c = v(t), where c would be vo. would it require to find time for both equations, if so, how would I go about doing this? Thank you

 

[HallsofIvy]

okay so I know that vo= vocos(f) + vosin(f) and they give you initial conditions for f=40 degrees and xfinal, or where the ball lands, is 194.55 meters and I know that speed is the magnitude of velocity but I keep getting 1.5867 but that is not correct. Any ideas where I went wrong?

I have no idea what kind of response you expect! in your original post you said nothing about "f= 40 degrees" nor did you say that the distance, R, was 194.55 meters. Where did you get that. And we cannot tell you where you "went wrong" if you won't tell us what you did!

 

[Deleted member 4993]

I have no idea what kind of response you expect! in your original post you said nothing about "f= 40 degrees" nor did you say that the distance, R, was 194.55 meters. Where did you get that. And we cannot tell you where you "went wrong" if you won't tell us what you did!

That was my doing!! When the problem was first posted, it came up with broken images. So put in generic variables to start the problem.

 

[jenquin12]

I explained my procedure in the above message. Thank you

 

[Deleted member 4993]

A golf ball is hit from ground level at an angle of 40° (Θ) degrees above ground level. The terrain is level. The ball lands 194.55 (R) meters away.What was the initial speed of the golf ball?
For that same initial speed, find two angles of departure for which the ball will travel to a distance of 194.55 meters. ( Answer in degrees)

No idea how to start this!! help!

Assuming initial speed = V

R = V²*sin(2Θ)/g.................Solve for V

 

[jenquin12]

thank you very much