Body thrown at an angle to the horizontal

[rbh]

Homework Statement

In the picture below you can see part of trajectory of a body thrown with an angle to the horizont. At point A the speed was 20m/s. How much time did it take for the body to move from point A to B?

Relevant Equations

V = Vo + at

The answer says t = Vb - Va/g. Shouldn't it be Va + V(from highest point to point b)/g ?

 

[haruspex]

The question as quoted does not provide enough information. It says nothing about a speed or velocity at B, yet you say that forms part of the official answer.
Please be sure to quote the question exactly as given to you.

The given answer makes sense provided:
- you mean (Vb-Va)/g
- Va and Vb are the vertical components of velocity at A and B
But you quote the speed at A, which would include the horizontal component.

 

[rbh]

The question as quoted does not provide enough information. It says nothing about a speed or velocity at B, yet you say that forms part of the official answer.
Please be sure to quote the question exactly as given to you.

The given answer makes sense provided:
- you mean (Vb-Va)/g
- Va and Vb are the vertical components of velocity at A and B
But you quote the speed at A, which would include the horizontal component.

The question is complete, I believe the author wanted to make this question a little bit more harder and trick into thinking there is a numerical answer since there was a second right answer with data which weren't mentioned in the problem.
Yes I was talking about y components. Could you explain why it is right? My thinking why t = (Vay + Vby)/g :
at the maximum point of the trajectory y velocity component becomes 0 and the projectile reaches speed Vby twice.

 

[Steve4Physics]

Yes I was talking about y components. Could you explain why it is right? My thinking why t = (Vay + Vby)/g :
at the maximum point of the trajectory y velocity component becomes 0 and the projectile reaches speed Vby twice.

Can I chip in? Maybe the problem is this: vay, vby and g are vectors. If you substitute actual values, the values for vby and for g will be negative (assuming upwards is positive). Try this:

Q1. Write an expression for the time it takes to go from A to the highest point. Check this will give a positive value.

Q2. Write an expression for the time is takes to go from the highest point to B. Check this will give a positive value.

Q3. If you add these two times, what do you get?

Q4. Can you see how to get the same answer simply by considering initial and final vertical velocities (without working out the two separate times)?

 

[haruspex]

I believe the author wanted to make this question a little bit more harder

It is nonsensical to leave the solver to choose which variables are to be taken as given, whether as numbers or variable names. Instead of assuming the vertical velocities at A and B are known, you could assume the time is known, then just answer "t".
There has to be something in the problem statement that tells you which unknowns are allowed in the answer.

vay, vby and g are vectors.

Having specified y components, I see no advantage in taking them as vectors. Perhaps you meant that v_A and v_B are vectors, so $\vec v_B-\vec v_A$ would be a vertical vector? But that still doesn't let us divide by vector $\vec g$.

 

[Steve4Physics]

Having specified y components, I see no advantage in taking them as vectors. Perhaps you meant that v_A and v_B are vectors, so $\vec v_B-\vec v_A$ would be a vertical vector? But that still doesn't let us divide by vector $\vec g$.

Yes. I could have expressed that better.

@rbh's formula "t = (Vay + Vby)/g" is inccorect and this looks like a result of dealing with signs/directions incorrectly. I wanted point @rbh in this direction [humour intended].