Time of flight of projectile from launch to landing
The formula below is the time of flight ( time of whole journey from launch(0,0) to landing (×,y) ) of a projectile whose initial vertical position is above the point of impact.
I am trying to understand how the right side of the equation is derived. For instance, how do I come up with 2gy$_0$ ?
$frac{d}{v.cos(theta)}$ = $frac{v.sin(theta)+sqrt{left(v.sin(theta)right)^2 +2gy_0}}{g}$
Where
g = gravitational acceleration
y$_0$ = initial vertical position (h)
d = entire horizontal distance or range of the flight from launch to landing
v = velocity
$theta$ = initial launch angle
Thanks
projectile-motion
asked Nov 27 '18 at 21:59
EdvilleEdville
596
add a comment |
The formula below is the time of flight ( time of whole journey from launch(0,0) to landing (×,y) ) of a projectile whose initial vertical position is above the point of impact.
I am trying to understand how the right side of the equation is derived. For instance, how do I come up with 2gy$_0$ ?
$frac{d}{v.cos(theta)}$ = $frac{v.sin(theta)+sqrt{left(v.sin(theta)right)^2 +2gy_0}}{g}$
Where
g = gravitational acceleration
y$_0$ = initial vertical position (h)
d = entire horizontal distance or range of the flight from launch to landing
v = velocity
$theta$ = initial launch angle
Thanks
projectile-motion
asked Nov 27 '18 at 21:59
EdvilleEdville
596
You can find the proof in any good physics book. Also refer to Projectile_motion.
– gimusi
Nov 27 '18 at 22:04
This was actually from wikipedia. It didn't show how the right hand side is derived, and I am unable to find an online resource for this particular derivation.
– Edville
Nov 27 '18 at 22:09
Ah ok! I've added a hint to find the solution.
– gimusi
Nov 27 '18 at 22:14
add a comment |
The formula below is the time of flight ( time of whole journey from launch(0,0) to landing (×,y) ) of a projectile whose initial vertical position is above the point of impact.
I am trying to understand how the right side of the equation is derived. For instance, how do I come up with 2gy$_0$ ?
$frac{d}{v.cos(theta)}$ = $frac{v.sin(theta)+sqrt{left(v.sin(theta)right)^2 +2gy_0}}{g}$
Where
g = gravitational acceleration
y$_0$ = initial vertical position (h)
d = entire horizontal distance or range of the flight from launch to landing
v = velocity
$theta$ = initial launch angle
Thanks
projectile-motion
asked Nov 27 '18 at 21:59
EdvilleEdville
596
The formula below is the time of flight ( time of whole journey from launch(0,0) to landing (×,y) ) of a projectile whose initial vertical position is above the point of impact.
I am trying to understand how the right side of the equation is derived. For instance, how do I come up with 2gy$_0$ ?
$frac{d}{v.cos(theta)}$ = $frac{v.sin(theta)+sqrt{left(v.sin(theta)right)^2 +2gy_0}}{g}$
Where
g = gravitational acceleration
y$_0$ = initial vertical position (h)
d = entire horizontal distance or range of the flight from launch to landing
v = velocity
$theta$ = initial launch angle
Thanks
projectile-motion
projectile-motion
asked Nov 27 '18 at 21:59
EdvilleEdville
596
asked Nov 27 '18 at 21:59
EdvilleEdville
596
asked Nov 27 '18 at 21:59
EdvilleEdville
596
asked Nov 27 '18 at 21:59
EdvilleEdville
596
asked Nov 27 '18 at 21:59
EdvilleEdville
596
596
You can find the proof in any good physics book. Also refer to Projectile_motion.
– gimusi
Nov 27 '18 at 22:04
This was actually from wikipedia. It didn't show how the right hand side is derived, and I am unable to find an online resource for this particular derivation.
– Edville
Nov 27 '18 at 22:09
Ah ok! I've added a hint to find the solution.
– gimusi
Nov 27 '18 at 22:14
add a comment |
You can find the proof in any good physics book. Also refer to Projectile_motion.
– gimusi
Nov 27 '18 at 22:04
This was actually from wikipedia. It didn't show how the right hand side is derived, and I am unable to find an online resource for this particular derivation.
– Edville
Nov 27 '18 at 22:09
Ah ok! I've added a hint to find the solution.
– gimusi
Nov 27 '18 at 22:14
You can find the proof in any good physics book. Also refer to Projectile_motion.
– gimusi
Nov 27 '18 at 22:04
You can find the proof in any good physics book. Also refer to Projectile_motion.
– gimusi
Nov 27 '18 at 22:04
This was actually from wikipedia. It didn't show how the right hand side is derived, and I am unable to find an online resource for this particular derivation.
– Edville
Nov 27 '18 at 22:09
This was actually from wikipedia. It didn't show how the right hand side is derived, and I am unable to find an online resource for this particular derivation.
– Edville
Nov 27 '18 at 22:09
Ah ok! I've added a hint to find the solution.
– gimusi
Nov 27 '18 at 22:14
Ah ok! I've added a hint to find the solution.
– gimusi
Nov 27 '18 at 22:14
add a comment |
1 Answer
1
active
oldest
votes
HINT
Let consider at first the equation of motion in vertical direction that is
- $y(t)=h+v_0 sintheta cdot t-frac12 g t^2$
then by the condition $y(t)=0$ find the time of landing $t_{L}$.
Finally use that to find x of landing by
- $d=x(t_{L})=v_0 cos theta cdot t_{L}$
answered Nov 27 '18 at 22:13
gimusigimusi
1
Thanks a lot for this. I can now see that it involves the use of quadratic formula as shown here khanacademy.org/science/physics/two-dimensional-motion/modal/v/…
– Edville
Nov 28 '18 at 11:45
Yes exactly! Let me know if you need some more information. Bye
– gimusi
Nov 28 '18 at 13:26
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
HINT
Let consider at first the equation of motion in vertical direction that is
- $y(t)=h+v_0 sintheta cdot t-frac12 g t^2$
then by the condition $y(t)=0$ find the time of landing $t_{L}$.
Finally use that to find x of landing by
- $d=x(t_{L})=v_0 cos theta cdot t_{L}$
answered Nov 27 '18 at 22:13
gimusigimusi
1
Thanks a lot for this. I can now see that it involves the use of quadratic formula as shown here khanacademy.org/science/physics/two-dimensional-motion/modal/v/…
– Edville
Nov 28 '18 at 11:45
Yes exactly! Let me know if you need some more information. Bye
– gimusi
Nov 28 '18 at 13:26
add a comment |
HINT
Let consider at first the equation of motion in vertical direction that is
- $y(t)=h+v_0 sintheta cdot t-frac12 g t^2$
then by the condition $y(t)=0$ find the time of landing $t_{L}$.
Finally use that to find x of landing by
- $d=x(t_{L})=v_0 cos theta cdot t_{L}$
answered Nov 27 '18 at 22:13
gimusigimusi
1
Thanks a lot for this. I can now see that it involves the use of quadratic formula as shown here khanacademy.org/science/physics/two-dimensional-motion/modal/v/…
– Edville
Nov 28 '18 at 11:45
Yes exactly! Let me know if you need some more information. Bye
– gimusi
Nov 28 '18 at 13:26
add a comment |
HINT
Let consider at first the equation of motion in vertical direction that is
- $y(t)=h+v_0 sintheta cdot t-frac12 g t^2$
then by the condition $y(t)=0$ find the time of landing $t_{L}$.
Finally use that to find x of landing by
- $d=x(t_{L})=v_0 cos theta cdot t_{L}$
answered Nov 27 '18 at 22:13
gimusigimusi
1
HINT
Let consider at first the equation of motion in vertical direction that is
- $y(t)=h+v_0 sintheta cdot t-frac12 g t^2$
then by the condition $y(t)=0$ find the time of landing $t_{L}$.
Finally use that to find x of landing by
- $d=x(t_{L})=v_0 cos theta cdot t_{L}$
answered Nov 27 '18 at 22:13
gimusigimusi
1
answered Nov 27 '18 at 22:13
gimusigimusi
1
answered Nov 27 '18 at 22:13
gimusigimusi
1
answered Nov 27 '18 at 22:13
gimusigimusi
1
1
Thanks a lot for this. I can now see that it involves the use of quadratic formula as shown here khanacademy.org/science/physics/two-dimensional-motion/modal/v/…
– Edville
Nov 28 '18 at 11:45
Yes exactly! Let me know if you need some more information. Bye
– gimusi
Nov 28 '18 at 13:26
add a comment |
Thanks a lot for this. I can now see that it involves the use of quadratic formula as shown here khanacademy.org/science/physics/two-dimensional-motion/modal/v/…
– Edville
Nov 28 '18 at 11:45
Yes exactly! Let me know if you need some more information. Bye
– gimusi
Nov 28 '18 at 13:26
Thanks a lot for this. I can now see that it involves the use of quadratic formula as shown here khanacademy.org/science/physics/two-dimensional-motion/modal/v/…
– Edville
Nov 28 '18 at 11:45
Thanks a lot for this. I can now see that it involves the use of quadratic formula as shown here khanacademy.org/science/physics/two-dimensional-motion/modal/v/…
– Edville
Nov 28 '18 at 11:45
Yes exactly! Let me know if you need some more information. Bye
– gimusi
Nov 28 '18 at 13:26
Yes exactly! Let me know if you need some more information. Bye
– gimusi
Nov 28 '18 at 13:26
add a comment |
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You can find the proof in any good physics book. Also refer to Projectile_motion.
– gimusi
Nov 27 '18 at 22:04
This was actually from wikipedia. It didn't show how the right hand side is derived, and I am unable to find an online resource for this particular derivation.
– Edville
Nov 27 '18 at 22:09
Ah ok! I've added a hint to find the solution.
– gimusi
Nov 27 '18 at 22:14