Pendulum - Maximum Angle in 2nd half of swing

[bigsmile]

A simple pendulum, 2.0 m in length, is released by a push when the support string is at an angle of 25° from the vertical. If the initial speed of the suspended mass is 1.2 m/s when at the release point, to what maximum angle will it move in the second half of its swing?

Let h1 be the elevation reached when potential energy = total energy transferred.
mg(h1) = mgL(1 - cos θ0) + 1/2 (mV²)

I am not sure if the approach is correct and how to proceed. My trig. background is not that strong at all.

Please help. I believe the answer is 30 degrees.

 

[mfb]

The approach is correct. You know everything apart from h1 (and m, but you can divide the equation by m to get rid of it), so you can solve that equation.
With h1 and L you can find the angle.
Alternatively, directly use the angle-dependent expression on the left side.

 

[bigsmile]

Thanks for you feedback. I assume h1 = L ( 1 - cos 25 degrees) . Is that right?
I have troubles to find the "V" . I don't think I should use the initial speed 1.2 m/s.
Since it is asking for max angle for 2nd half of swing. Beside theta 0, do I need to set up theta 1 , another angle for 2nd half of swing? Confused.
Need some guidance.

 

[mfb]

I assume h1 = L ( 1 - cos 25 degrees) . Is that right?

Instead of 25 degrees, you have to use the angle you want to calculate there.

I have troubles to find the "V" . I don't think I should use the initial speed 1.2 m/s.

Why not?