A block is launched up a frictionless 40° slope with an initial speed v and reaches a maximum vertical height h. The same block is launched up a frictionless 20° slope with the same initial speed v. On this slope, the block reaches a maximum vertical height of _____.

1. h
2. h/2
3. 2h
4. A height greater than h but less than 2h.
5. A height greater than hl2 but less than h.

The block would reach exactly the same height from the ground. It would travel a greater distance away from the source, but the height away from the earth would remain the same as you are giving it the same energy each time. Therefore, it will reach the same gravitation potential energy.

Another approach to look at it this is seeing it when the Block moves up the slope, its kinetic energy decreases and the potential energy increases. In both cases, the kinetic energy decreases by same amount, therefore the block rises to same height H.

Try to use the formula;

1/2MV2 = mgh

Where V = √(2gh)

I hope this helps

Cricetus Cricetus · Answer 2 · 2021-11-28T16:34:16+01:00

The block reaches a maximum vertical height of "h".

During maximum vertical elevation, mass's initial energy released (K.E) would indeed be transformed towards the potential energy.
Because the initial kinetic energy would be similar at each slope, this same potential energy as well as consequently the elevation would still be the similar one.

Thus the above response i.e., "option 1." is appropriate.

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