Answer:
h=18.05 cm
Explanation:
Given that
m= 25 kg
K= 1300 N/m
x=26.4 cm
θ= 19.5 ∘
When the block just leave the spring then the speed of block = v m/s
From energy conservation



By putting the values


v=1.9 m/s
When block reach at the maximum height(h) position then the final speed of the block will be zero.
We know that

By putting the values

h=0.1805 m
h=18.05 cm
Answer: 1.6Hz
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The surrounding environment cool off is the answer
I think it false. Sorry if i'm wrong.
King Arthur's knights use a catapult to launch a rock from their vantage point on top of the castle wall, 14 m above the moat. The rock is launched at a speed of 27 m/s and an angle of 32degrees above the horizontal.