# Thread: [SOLVED] ideal projectile motion (vectors) calc3

1. ## [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.

2. $\displaystyle v_0 = 840m/s$

$\displaystyle \alpha=60~degrees$

$\displaystyle 21km=21000m$

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

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

Now look the distance travelled horizontally is known as Downrange

Thus

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

Can you go ahead

3. Originally Posted by ADARSH
$\displaystyle v_0 = 840m/s$

$\displaystyle \alpha=60~degrees$

$\displaystyle 21km=21000m$

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

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

Now look the distance travelled horizontally is known as Downrange

Thus

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

Can you go ahead

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?

4. 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

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

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

-similarly $\displaystyle vcos(\theta)$ 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 $\displaystyle V_x$ ONLY)

-We can use the general formula that

$\displaystyle Velocity~\times ~time=~distance$

ie;

$\displaystyle V_x ~\times t = 21000m$

Watch the diagram and this link