Projectile motion of two balls off a cliff

[tellmesomething]

Homework Statement

1. Two balls are projected from the top of a cliff with equal initial speed u. One starts at angle θ
above the horizontal while the other starts at angle θ below. Difference in their ranges is

Relevant Equations

Non

This is just the object which is thrown theta angle below rhe horizontal. I fail to understand how to get the individual range of this. I tried but I am getting zero, can someone point out what is wrong in my method

So $R=ucos\theta*t$
$\frac{h}{R}=tan\theta$
Therefore $h=Rtan\theta$
We know that $H=usin\theta*t+\frac{1}{2}gt^2$
So $Rtan\theta=\frac{R}{ucos\theta}[usin\theta +\frac{1}{2}g\frac{R}{ucos\theta}]$
As you can see we get
$usin\theta=usin\theta+\frac{1}{2}g\frac{R}{ucos\theta}$

So R=0

What is wrong here

 

[PeroK]

Homework Statement: 1. Two balls are projected from the top of a cliff with equal initial speed u. One starts at angle θ
above the horizontal while the other starts at angle θ below. Difference in their ranges is
Relevant Equations: Non

View attachment 353821
This is just the object which is thrown theta angle below rhe horizontal. I fail to understand how to get the individual range of this. I tried but I am getting zero, can someone point out what is wrong in my method

So $R=ucos\theta*t$
$\frac{h}{R}=tan\theta$
Therefore $h=Rtan\theta$
We know that $H=usin\theta*t+\frac{1}{2}gt^2$
So $Rtan\theta=\frac{R}{ucos\theta}[usin\theta +\frac{1}{2}g\frac{R}{ucos\theta}]$
As you can see we get
$usin\theta=usin\theta+\frac{1}{2}g\frac{R}{ucos\theta}$

So R=0

What is wrong here

The projectile moves in a parabola. The angle $\theta$ is only the initial angle.

 

[erobz]

Homework Statement: 1. Two balls are projected from the top of a cliff with equal initial speed u. One starts at angle θ
above the horizontal while the other starts at angle θ below. Difference in their ranges is
Relevant Equations: Non

View attachment 353821
This is just the object which is thrown theta angle below rhe horizontal. I fail to understand how to get the individual range of this. I tried but I am getting zero, can someone point out what is wrong in my method

So $R=ucos\theta*t$
$\frac{h}{R}=tan\theta$
Therefore $h=Rtan\theta$
We know that $H=usin\theta*t+\frac{1}{2}gt^2$
So $Rtan\theta=\frac{R}{ucos\theta}[usin\theta +\frac{1}{2}g\frac{R}{ucos\theta}]$
As you can see we get
$usin\theta=usin\theta+\frac{1}{2}g\frac{R}{ucos\theta}$

So R=0

What is wrong here

Your diagram doesn't seem to make sense for the problems neither do your kinematic models?

I would start by writing the range of each projectile after I drew two diagrams for comparison. Begin with the projectile that is projected above horizontal. Solve the system of kinematic equations for it, what do you get for its range?

P.S. after I did that, I realized I could satisfy both scenarios with the same equation by seeing the sign properties of the trig functions with regards to $\theta$ and $-\theta$ arguments.

 

[Steve4Physics]

@tellmesomething, avoid using different symbols (h and H) for the same quantity.

It looks like you are sticking numbers into equations and doing the algebra without fully understanding the physics – dangerous! Here’s a possible approach:

1) For the upwards throw, note that the initial vertical component of velocity is usin⁡θ upwards. And for the downwards throw, it is usin⁡θ downwards.

2) Use the above to find the time-of-flight for each throw. The times are different. (The times aren’t affected by the horizontal component of motion.)

3) The horizontal component of velocity for both throws is ucos⁡θ so, knowing the times, you can now find the two ranges.

Edit: It's worth noting that there is a quick method to solve the problem. This depends on the fact that the time-difference between the throws can easily be found - if you can spot the 'trick'.