Simple velocity question

A ball thrown directly upwards rises to a highest point and falls back into the throwers hand; what is the average velocity?

By $\Delta$Distance=Average Velocity x $\Delta$time
the Va would be O since $\Delta$distance is 0, which is what the answer says.

But then if I was to go by Average velocity = (Final velocity + Initial velocity)/2
wouldn't it not be zero since there had to be some final velocity on its way down?

If the final velocity is +v then the initial velocity will be -v (opposite direction)
Final + initial = (+v + (- v) )/2 = 0

thanks! Does the negative v for the opposite direction also apply when finding average velocity vf+vi/2 ?

Yes, but you need to be careful with this.
I should have pointed out from your 1st post that it should be
distance = average speed x time
displacement = average velocity x time

distance and speed are scalars
displacement and velocity are vectors

If you throw a ball up to a height of, say 2m, and catch it the total displacement is zero. The distance travelled is 4m.

The average speed will be 4m / time taken
The average velocity will be 0m / time taken


If you travel round a circle of radius r, your displacement is 0. Your distance travelled is 2πr. Your average velocity and average speed will consequently be different.
It's an important distinction in physics.

Average speed is defined as total distance travelled divided by time taken.
Average velocity is (v₁ + v₂)/2 only for uniform acceleration in a straight line. In general it's total displacement / time taken.

Thank you very much!
One last question, given the current information, how should I proceed with finding x?

Could you post the complete question, please.

Well, it wasn't a really a question; they just used values already known to calculate the range.

I just wanted to know that given a question with a negative angle( I can't find a proper one), how I should proceed with finding the range.


If the angle had been positive, I'd find the time it took for the object to reach its max vertical height and then multiply that by 2 to get the time it took for the object to reach the endpoint horizontally
The initial velocity would have been given and I'd take the horizontal component of that.
Id then use s=ut to get the range


But with a negative angle I don't know how to proceed.

The distance x is 25t where t is the time of flight (horizontal motion)
The distance y = ½gt² (from s = ut + ½gt² on the vertical motion)

You also know y/x = tan 35
from this you can find t
From t you can find x and y