Finding Range of the Projectile

[Abhi13]

Hi All,

I have a question regarding finding the range of the projectile given the projectile crosses points P1(a,b) and P2(b,a).

I know any point on the projectile curve satisfies following equation:
y = x.tan \{theta} (1- (range/x))

therefore, the only thing remains to be found out is tan \{theta}.

Can anyone please direct me in finding \theta given these two point- P1(a,b) and P2(b,a).

Thanks,
Abhi

 

[cragar]

By range do you mean from the lowest y value to the highest y value ,
or the difference between the x values .
Like when i throw a golf ball the range it traveled over the ground . In that sense

 

[Abhi13]

By range do you mean from the lowest y value to the highest y value ,
or the difference between the x values .
Like when i throw a golf ball the range it traveled over the ground . In that sense

By range i mean the distance it traveled in the horizontal direction. Thats the difference between the values of x from where it took off to where it will land..

 

[cragar]

I'm confused by your equation you have .
is it y=x*tan(q)(1-r/x)

 

[PJC01]

Hello Abhi,

Thank you for your question. Finding the range of a projectile can be a complex problem, but there are a few key steps you can take to help determine the angle (\theta) needed to reach the points P1(a,b) and P2(b,a).

First, it is important to understand the definition of range in this context. Range is defined as the horizontal distance traveled by the projectile before it hits the ground. This means that the y-value of the projectile must be equal to 0 at the point of impact.

Using the equation you provided, y = x.tan\{theta} (1-(range/x)), we can substitute in the values for P1 and P2 to get two equations:

P1: b = a.tan\{theta} (1-(range/a))
P2: a = b.tan\{theta} (1-(range/b))

Since we are trying to find the angle \theta, we can rearrange these equations to solve for it. This can be done by isolating the term tan\{theta} on one side of the equation:

P1: tan\{theta} = b/(a(1-(range/a)))
P2: tan\{theta} = a/(b(1-(range/b)))

Now we have two equations in terms of tan\{theta}. We can use basic algebra to set these two equations equal to each other, since they both equal tan\{theta}:

b/(a(1-(range/a))) = a/(b(1-(range/b)))

Next, we can cross-multiply and solve for range:

b(b(1-(range/b))) = a(a(1-(range/a)))
b(1-(range/b)) = a(1-(range/a))
b-b(range/b) = a-a(range/a)
b = a - a(range/a) + b(range/b)
a+b(range/b) = a-b(range/a)
a = a - b(range/a) - b(range/b)
b(range/a) + b(range/b) = 0
b(range/a + range/b) = 0
range(a+b) = 0
range = 0/(a+b)

We can see that in order for the range to be equal to 0, the denominator (a+b) must also equal 0. This means that a+b = 0, or a = -b. This condition must be