Free fall question

[CellCoree]

A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 6.35m/s . Air resistance may be ignored, so the water balloon is in free fall after it leaves the thrower's hand.


A.) What is its speed after falling for a time 1.65s ? Take the free fall acceleration to be g=9.80m/s^2 .

my work:
using formula: v(t) = v(0) + at
v(t) = (6.35) + (-9.8)(1.65) = -9.82

B.) How far does it fall in a time of 1.65s ? Take the free fall acceleration to be 9.80 m/s^2.

my work: x(t) = x(0) + v(0)t + 1/2at^2
x(0) = (6.35)(1.65) + 1/2(-9.8)(1.65)^2
x = 2.86

C.) What is the magnitude of its velocity after falling a distance 10.9m ? Take the free fall acceleration to be g=9.80 m/s^2.

my work:
v^2 = v(0)^2 + 2a(x-x(0))
v^2 = +2(-9.8)(10.9)
v= 14.616

ok, i got them all wrong. what am i doing wrong? i tried everything that i know and can't get it right, please help

 

[tyco05]

From what I can tell, you have the right workings, but you are using a wrong value for the acceleration - - - - Remember that in this case, the acceleration is acting in the same direction as the velocity.

This should fix up all of your problems

 

[PureEnergy]

Since the ball is thrown downward it has a velocity of -6.35 and not 6.35, be sure to always check the direction. For the last part, don't forget that you begin with a velocity of -6.35 m/s.

 

[tyco05]

Yes, or alternatively, you could use different values for your velocity

it's all about assigning direction

if you use -ve for gravity, then you must use -ve for velocity,
or you could make them both +ve.

 

[CellCoree]

Yes, or alternatively, you could use different values for your velocity

it's all about assigning direction

if you use -ve for gravity, then you must use -ve for velocity,
or you could make them both +ve.

what is -ve?

and why is the answer for problem A.) a positive answer? after inserting -6.35, i get a negative answer. so... why is the answer positive instead of negative?


"For the last part, don't forget that you begin with a velocity of -6.35 m/s"

are you talking about problem C.)?

if so would it be v^2 = (-6.35)^2 + 2(-9.8)(10.9)?

 

[tyco05]

-ve = negative
+ve = positive

The answer can be either negative or positive, depending on how you define your directions. It doesn't matter (whether the direction of the motion is positive or negative) as long as you are consistent with your choice.
That is, if you are saying that your acceleration is negative (towards the ground) and the object is moving towards the ground, then the object's velocity also has to be negative. (as you have assigned negative motion to be motion towards the ground.)

You could just as easily assign motion towards the ground to be positive motion.
Just remember that your velocity is in the same direction as your acceleration in this problem.