Is the Final Velocity of a Projectile Always Greater Than the Initial Velocity?

[jonniechung]

Homework Statement

Hi, I am just wondering if someone can check if this is the correct answer for this question. Thanks in advance

1 . A projectile is launched with a speed of 48m/s at an angle of 34 degrees BELOW the horizontal. Calculate the speed an direction of the projectile 2.5 seconds after launch. Ignore friction.

Homework Equations

x/y component . addition of vectors. v=vo+at

The Attempt at a Solution

This is my answer : 40 m/s at 3.3 degress South of East

I find it surprising that the final velocity of the projectile is less then the initial (which is 48m/s).


Here it is:

48 sin 34 = -26.8m/s
48 cos 34 = 39.8

v = vo +at

v = -26.8 + (9.8)(2.5) <--- + acceleration because motion is towards center of Earth

v = -2.3

From here I used addition of vectors in x/y component

x = 39.8
y = -2.2

After 2.5 seconds

Answer: vf = 40 m/s at an angle of 3.3 degrees <---- tan theta


Shouldn't the vf of the projectile (after 2.5 seconds) be greater then the vo? Since it has a + 9.8 m/s ^ 2 .

Thank you

 

[Delphi51]

v = -26.8 + (9.8)(2.5) <--- + acceleration because motion is towards center of Earth

You need a minus sign on the 9.8. The 26.8 and the 9.8*2.5 are both downward.
For the direction, you'll get an angle below horizontal - no "South of East"

 

[jonniechung]

You need a minus sign on the 9.8. The 26.8 and the 9.8*2.5 are both downward.
For the direction, you'll get an angle below horizontal - no "South of East"

Thank you very much. The correct answer should be 65 m/s , 55 degrees below horizontal ?

 

[Delphi51]

I got 64 m/s and 51 degrees, but I used g = 9.81.
Might be worth checking the angle again - that's quite a difference.
Did you have 49.7 for the vertical velocity?

 

[jonniechung]

Nope, i got 51.4 m/s...

But... it was 52 degrees. I don't know why i put down 55 degrees lol

Thanks