Projectile motion using energy method

[pringless]

i)A rocket is launched at an angle of 70.9 degrees from the horizontal from an altitude of 5.4m with a speed of 6.3m/s. Use energy methods to find its speed when its altitude is 2.7 m.

I got a correct answer of 9.6234 m/s. I need help for the next part.

ii)Find the y component of the velocity when the rocket's altitude is 2.7m.

i don't quite what velocity i should use? i tried using the velocity i got for the first part but the answer is incorrect.

 

[chroot]

The vertical component of the initial velocity is 6.3 \sin 70.9^o\ m/s. (The horizontal component is 6.3 \cos 70.9^o\ m/s.)

Think about it this way: the vertical motion and the horizontal motion are independent. If you drop a rock off the edge of a cliff (so it has zero horizontal velocity), you know how to calculate how long it'll take to hit the ground below. If you throw the rock off the cliff (so it has some horizontal velocity), you know that it does not affect how long it takes to hit the ground below.

So, you know the initial vertical velocity, and you know how far the rocket fell. You can use that calculate its final vertical velocity. Its horizontal velocity is constant (and irrelevant).

Using energy arguments:

\frac{1}{2} m v_f^2 = \frac{1}{2} m v_i^2 + m g h

v_f = \sqrt{v_i^2 + 2 g h}

Plug in the initial vertical velocity, 6.3 \sin 70.9^o\ m/s -- you should find the final vertical velocity is 9.4 m/s.

Does this answer agree with your previous answer? After it has fallen to 2.7 m altitude, the rocket has a vertical velocity of 9.4 m/s, and the same horizontal velocity it started with: 6.3 \cos 70.9^o\ m/s. You can add those components together to find the total velocity:

\begin{align*}<br /> v &amp;= \sqrt{v_x^2 + v_y^2}<br /> &amp;= \sqrt{(6.3 \cos 70.9^o)^2 + 9.4^2}<br /> &amp;= 9.624\ m/s<br /> \end{align*}<br />

The answer checks. Does this make sense?

- Warren

 

[pringless]

Hey, Thanks a lot Warren. I understand now.