# Distance of projectile - only speed given

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1. May 10, 2017

### SpiraRoam

1. The problem statement, all variables and given/known data
Andy Roberts the former West Indian Cricket player and Fast Bowler, bowled his fastest delivery in 1975 at 159.5 km/h. Neglecting air resistance, calculate the maximum distance he could have thrown the ball at this speed (on earth!) had he been able to throw it:

i) Vertically upwards from the cricket field

ii) Horizontally across the cricket field.

2. Relevant equations
s=ut+1/2at^2
v=u + at (rearranged for a and t)
distance = speed x time

3. The attempt at a solution
I'm pretty confused with this problem - I'm used to a couple of values being given with equations of motion and projectiles. I was wondering whether to find the acceleration of the ball or to just assume that it's constant? The value given is a velocity with it having a direction (horizontal / vertical) making it a vector right?

I'm not sure whether the velocity stays constant throughout either? If it's being thrown the maximum distance then it's bound to lose velocity as time progresses. How am I to calculate this for the final velocity and how would this impact on the time?

Converting the 159.5 km/h to metres per second gives 44.31 m/s. This will be the initial velocity but how to find the acceleration and time values? I'm sure gravity comes into it for the horizontal aswell as vertical directions and I imagine the ball won't be airborne for more than 3-5 seconds or so but how to calculate the certainty?

I was looking into using a speed - time graph and finding distance from the area underneath but again there aren't enough values given and I don't know how to derive them.

Thanks

Last edited: May 10, 2017
2. May 10, 2017

### Staff: Mentor

The problem is oddly worded, but I interpret it as: What's the maximum height the ball could be thrown? and: What's the maximum distance across the field the ball could be thrown?

If the velocity stayed constant the ball would just keep going higher and higher!

Hint: This is a projectile motion problem.

3. May 10, 2017

### SpiraRoam

I need to find the time in seconds for the vertical trajectory and then I can multiply that by the 44.31 m/s for the horizontal distance.

I'm stuck with how to find the vertical time or distance though. I'm sure it's derived from s=1/2at^2 and for time this would be rearranged to:

t^2= 2s/a and then square rooting.

Would there be an acceleration for the vertical path with gravity acting down on the ball? In most of the example problems I'm reading about there is always at least one more value given.

4. May 10, 2017

### CWatters

The question is very badly worded....

a) If he throws it straight up the ball will start decelerating at "g" as soon as it leaves his hand. To calculate how high it will go you can either apply conservation of energy or use the equations of motion with constant acceleration. It's not clear what "maximum distance" means. Is it the vertical height achieved? Double that? or zero because it lands back where it starts.

b) If he throws it horizontally you need to know how tall he is in order to calculate the time it take to fall to the ground. That's not given so you cannot answer this question without making an assumption about his height.

Lets assume there is a part c) that asks how far he could throw the ball if given a choice of launch angles..

It can be shown that the max range is achieved with a 45 degree launch angle. This is then a projectile motion problem. You need to calculate the horizontal and vertical components of the launch velocity. The vertical component and the equations of motion give you the time of flight. That and the horizontal component of velocity gives you the range.

5. May 10, 2017