Projectile Motion ball is launched

[Tanya Sharma]

Homework Statement

A ball is launched with initial speed $2\sqrt{}gh$ such that it just clears two walls of height 'h' separated by a distance 2h .What is the time taken by the ball between the two walls ?

Ans: $2\sqrt{\frac{h}{g}}$

Homework Equations

The Attempt at a Solution

Considering the origin to be at the launching point ,θ to be angle at which the ball is projected and 'x' to be distance between the origin and the first wall .

$$h = xtanθ-\frac{1}{2}\frac{gx^2}{u^2cos^2θ}$$

$$h = (x+2h)tanθ-\frac{1}{2}\frac{g(x+2h)^2}{u^2cos^2θ}$$

Solving the above two equations , we get x = h(2sin2θ-1) .

Now sure what to do next . I would be grateful if somebody could help me with the problem.

 

[ehild]

$2\sqrt{gh}$ can not be angle. Is it initial speed?

 

[Tanya Sharma]

Sorry ... You are right . It is initial speed . I have corrected it.

 

[ehild]

The answer is also wrong.

Anyway: you need time, so use the equations which include time.

 

[Tanya Sharma]

I have corrected the answer . Will reattempt this problem .

Thank you very much .

 

[voko]

You do not care about the launching point and the initial launch angle. You can trivially the speed of the ball as it reaches the first wall, which reduces the problem to motion between the walls.

 

[lep11]

The answer is also wrong.

Yes, shouldn't it be √(h/g)?

 

[ehild]

Yes, shouldn't it be √(h/g)?

The OP has been corrected. The answer 2√(h/g) is right.

 

[DocZaius]

You do not care about the launching point and the initial launch angle. You can trivially the speed of the ball as it reaches the first wall, which reduces the problem to motion between the walls.

Can you elaborate on the triviality of the speed of the ball as it reaches the first wall? I'm guessing it falls out of the symmetry of the problem, but I must say that it isn't obvious to me.

 

[voko]

The initial speed is known;
so the initial kinetic energy known;
the height of the wall is known;
so the potential energy at the wall is known;
...
PROFIT

 

[DocZaius]

The initial speed is known;
so the initial kinetic energy known;
the height of the wall is known;
so the potential energy at the wall is known;
...
PROFIT

Hah, fair enough, I was completely ignoring energy considerations.