# [SOLVED] ideal projectile motion (vectors) calc3

• Feb 15th 2009, 08:53 AM
Legendsn3verdie
[SOLVED] ideal projectile motion (vectors) calc3
travel time:
a projectile is fired at a speed of 840 m/sec at an angle of 60 degree. how long will it take to get 21km downrange?

here is what i did an its no where close to right. the answer was 50seconds i got no ideal how to get that. here is what i attempted to do.
http://i176.photobucket.com/albums/w...titled-173.jpg
• Feb 15th 2009, 09:09 AM
$
v_0 = 840m/s
$

$
\alpha=60~degrees
$

$21km=21000m$

Now Lets take horizontal component of velocity which remains constant
(as the gravity is vertical)
So
$
V_x= Vcos(60) = \frac{v}{2}
$

Thus in time t the distance travelled
$= \frac{v}{2} \times t$

Now look the distance travelled horizontally is known as Downrange

Thus

$\frac{840}{2} \times t= 21000$

• Feb 15th 2009, 09:14 AM
Legendsn3verdie
Quote:

$
v_0 = 840m/s
$

$
\alpha=60~degrees
$

$21km=21000m$

Now Lets take horizontal component of velocity which remains constant
(as the gravity is vertical)
So
$
V_x= Vcos(60) = \frac{v}{2}
$

Thus in time t the distance travelled
$= \frac{v}{2} \times t$

Now look the distance travelled horizontally is known as Downrange

Thus

$\frac{840}{2} \times t= 21000$

ok so 21000 * 2 = 42k / 840 = 50

i have no idea what equation you used to get that though..

the book gives a eqaution:

(2v sin(alpha)) /g = t

not sure what equation u used?
• Feb 15th 2009, 09:43 AM
Quote:

Originally Posted by Legendsn3verdie
ok so 21000 * 2 = 42k / 840 = 50

i have no idea what equation you used to get that though..

the book gives a eqaution:

not sure what equation u used?

-First of all You must know that velocity is in the form as

$
V^2= (vsin(\theta))^2+(vcos(\theta))^2
$

-Now The term $vsin(\theta)$ is the value of velocity in the y direction(vertical direction)
(use trignometry)

-similarly $vcos(\theta)$ is the value of velocity in the x(horizontal) direction

-Now the velocity in x direction remains constant and its value in y direction is affected by acceleration

- Since we were given
+the value of velocity
+ the value of alpha
+ the horizontal distance covered (distance travelled by $V_x$ ONLY)

-We can use the general formula that

$
Velocity~\times ~time=~distance
$

ie;

$
V_x ~\times t = 21000m
$

Watch the diagram and this link