How to work out time a ball was in the air.

Use your basic physics formulae. The horizontal movement of the ball has nothing to do with the time it spends in the air. The appropriate formula is s = ut + at^2. The easiest way is to measure the time from the apex to the ground, then double it to include the ascending phase. You have ‘s’, ‘u’ is zero at the apex, and ‘a’ is -g, or -9.8 ms^-2. From there, finding time is just solving the quadratic.

You can do the calculations yourself. And since I have practically answered your homework for you, do it again using the equation v^2 = u^2 + 2as just for practice. For your physics exams, make sure you can answer every question in at least two different ways. There is never only one correct method.

Or you can work it out by looking at the horizontal axis only and noticing when it left the ground and then touched it again. What is that distance? You know the horizontal velocity is 15 m/s. Distance = velocity x time. So time = distance / velocity.
Using this method the height is irrelevant.

@FireMadeFlesh
what is ut? Na, still not getting what your saying.

I don’t even get the question, couldn’t even begin to give the right answer as I have no idea what they are asking. What is the “horizontal component”?

I don’t have average speed, final velocity, time,

u is the initial velocity and t is the time at which this happens. The product of the two is zero since you are starting from the moment the ball reaches the apex when the velocity is zero.

Not trying to belittle, but why are you doing physics problems when you haven’t been taught about component velocities or vectors?

@worriedguy That will get you the answer, but it ignores the reason behind it and won’t be too helpful for similar questions in the future.

@Siblinings As @Shuttle128 said, ‘u’ is the initial velocity and ‘t’ is time. Average speed is not helpful, final velocity can be calculated from the given information, and so can time (which is what you are supposed to be doing). I’m not sure how to explain it any more clearly than I have.
The horizontal component is the part of the ball’s movement along the x axis. Movement in any given direction can be viewed as the sum of two vectors, in the x and y planes respectively. The x plane is irrelevant for calculating the time in the air, as the only factors affecting this exist in the y plane. The equation I gave you should be used to calculate the time the ball takes to fall back to earth from the apex of its motion, which will be exactly half of its time in the air.
To do physics properly, you need a means of drawing diagrams and writing equations properly. Unfortunately that limits my ability to explain that here.
A brilliant resource on all things physics is hyperphysics. Take a look at this link, particularly the time of flight section.

PS. I just realised I missed out part of the equation above – the true form is
s = ut + ½ at^2

There is not enough information given to solve the problem.

The horizontal velocity is how fast the ball is moving in a direction parallel to the ground. Ignoring friction with the air, this value remains constant.

What you need to know are the initial height from which the ball started its travel and the initial vertical velocity. You can take the initial ground coordinate to be 0, since this does not affect the shape of the curve.

Then you can graph the path of the ball by finding the x and y coordinates separately as functions of time. For a given time t the x coordinate is just 15t. The y coordinate can be found using the formula y = h + vt – 9.8t^2, where h is the initial height, v is the initial vertical velocity and 9.8 is the acceleration due to gravity in units of meters and seconds. The reason for the negative sign in front of 9.8 is that gravity causes downward acceleration.

I suppose that you could assume that v is zero and pick an arbitrary h to create the curve. The shape of the curve will be similar to an upside down version of the drawing that you linked to.

@LostInParadise The diagram in the link shows a peak height of 1.25m. There is enough information to solve it.

I was not sure how much information could be taken from the drawing, which is only supposed to suggest the shape.

Two corrections to what I said:
The formula for y should be:
y = h + vt – ½*9.8t^2

If you take v=0 then the pictue will look like the right half of the drawing linked to.

@Firemadeflesh The question was how long is the ball in the air? The graph plus the info that the horizontal speed is 15 m/s is enough. to get you the answer 2.5/15 = 0.1666 sec.

You can figure out gravity by looking at the slope of the line near the axis. If it is 45 deg you can figure the velocity at impact is the same as the horizontal velocity. Then you can work out the acceleration. V^2 = 2 a s Where s is vertical displacement

@worriedguy Yeah….you don’t actually need the horizontal velocity at all. The height at apex is all that is needed, and was provided. Let’s not confuse anyone here please.