# Find the time of movement

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1. Nov 6, 2016

### annalian

1. The problem statement, all variables and given/known data
A particle is thrown in the angle pi/3 radian with speed 10m/S. Find the time of movement if the maximal height is equal to the distance of fall.

2. Relevant equations
v^2-voy^2=2gh

3. The attempt at a solution
0-10^2*3/4=2*10*hmax
hmax=3.75
v0x*2t=3.75
v0*cos60*2t=3.75
10*0.5*2*t=3.75
t=0.375. The answer in my book is 1S.
I also want to know why whe I use the formula of maximal height v0^2sin^a/2g and maximal fallL=v0^2sin2a/g with the data from the problem they don't come out to be equal.

2. Nov 6, 2016

### Jamison Lahman

Can you explain what "time of movement" is? Is this the total time while the particle is in the air? What is the exact wording of the problem?

I am getting the same maximum height and using the equation $$h = \frac{1}{2}at^2$$ I am getting a total flight time of $\sqrt{3}$

Last edited: Nov 6, 2016
3. Nov 6, 2016

### haruspex

There seems to be too much information. That can be fixed by dropping the assumption that it lands at the same height it was launched at, but I still don't get the book answers.

4. Nov 6, 2016

### PhanthomJay

The angle of pi/3 radians might be measured from the vertical axis.

5. Nov 6, 2016

### haruspex

By dropping the assumption that it lands at the same height it was launched at, using pi/3 as angle to the horizontal I get a time of 1/4 s. Taking it as angle to the vertical makes it even less. To get a time of 1s I need a launch angle of acos(√2-1) to the horizontal, or about 65 degrees. That's using g=10m/s2.

6. Nov 6, 2016

### PhanthomJay

Poorly worded problem I can only assume it it is thrown at pi/3 radians with vertical on level ground. Vert comp of velocity thus 5 m/s. Rises to a height h then falls back vert distance h to ground, which is a 2nd assumption. Half sec up and half sec down. Problem should be tossed.

7. Nov 6, 2016

### haruspex

Right, so it was just a confusing way of saying it lands back at the height it was thrown from. Well deduced.