- #1
patv
- 7
- 0
[SOLVED] Arrow Being Launched Distance
An Arrow is launched with an initial velocity of 12.0m/s at an angle of 22.0 degrees with the horizontal, and is released at a height of 2.5m above the ground. How far away does it land on the ground?
d=vit + [tex]\frac{1}{2} [/tex]at2
vf2 = vi + 2ad
a =[tex]\frac{vf - vi} {t} [/tex] (I couldn't get this to display properly, but where it has sub in square brackets, it should be subscript.)
[tex]\sin{22}[/tex] =[tex]\frac{vy}{12}[/tex]
4.5m/s=vy
[tex]\cos{22}[/tex] = [ vx ] over 12 (I got an error the first time I tried this in latex, sorry)
vx = 11.14 m/s
After getting the velocity in the x and y direction, I attempted to find the height above the original shooting height the arrow would reach. I used a formula I cannot recall at the moment, and got 1m, but was informed that this was not correct.
Next, I will attempt to find the time the arrow will be in the air.
I tried to sub vt into the place of d in the equation below, and this gave me a time of 2.177 seconds. The substitute teacher said this was incorrect, but did not give a reason.
d=vit + [tex]\frac{1}{2}[/tex]at2
I know I'm supposed to find time first, but I really am not sure of how to get this. Can anyone provide assistance, or insight as to which formula to use, or where I am going wrong?
Thanks
Below is a diagram of how I think it would be drawn.
http://img60.imageshack.us/img60/9348/fw3tixcm6.th.jpg http://g.imageshack.us/thpix.php
Homework Statement
An Arrow is launched with an initial velocity of 12.0m/s at an angle of 22.0 degrees with the horizontal, and is released at a height of 2.5m above the ground. How far away does it land on the ground?
Homework Equations
d=vit + [tex]\frac{1}{2} [/tex]at2
vf2 = vi + 2ad
a =[tex]\frac{vf - vi} {t} [/tex] (I couldn't get this to display properly, but where it has sub in square brackets, it should be subscript.)
The Attempt at a Solution
[tex]\sin{22}[/tex] =[tex]\frac{vy}{12}[/tex]
4.5m/s=vy
[tex]\cos{22}[/tex] = [ vx ] over 12 (I got an error the first time I tried this in latex, sorry)
vx = 11.14 m/s
After getting the velocity in the x and y direction, I attempted to find the height above the original shooting height the arrow would reach. I used a formula I cannot recall at the moment, and got 1m, but was informed that this was not correct.
Next, I will attempt to find the time the arrow will be in the air.
I tried to sub vt into the place of d in the equation below, and this gave me a time of 2.177 seconds. The substitute teacher said this was incorrect, but did not give a reason.
d=vit + [tex]\frac{1}{2}[/tex]at2
I know I'm supposed to find time first, but I really am not sure of how to get this. Can anyone provide assistance, or insight as to which formula to use, or where I am going wrong?
Thanks
Below is a diagram of how I think it would be drawn.
http://img60.imageshack.us/img60/9348/fw3tixcm6.th.jpg http://g.imageshack.us/thpix.php
Last edited by a moderator: