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
[tex]\dfrac{1}{2}Kx^2=\dfrac{1}{2}mv^2[/tex]
[tex]Kx^2=mv^2[/tex]
[tex]v=\sqrt{\dfrac{kx^2}{m}}[/tex]
By putting the values
[tex]v=\sqrt{\dfrac{kx^2}{m}}[/tex]
[tex]v=\sqrt{\dfrac{1300\times 0.264^2}{25}}[/tex]
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
[tex]V_f^2=V_i^2-2gh[/tex]
By putting the values
[tex]0^2=1.9^2-2\times 10\times h[/tex]
h=0.1805 m
h=18.05 cm
