Answer:
a) 30.32 m/s
b) 46.855 m
Explanation:
F = Force
g = Acceleration due to gravity = 9.81 m/s²
s = Displacement
u = Initial velocity
v = Final velocity
Work done
[tex]W=Fs\\\Rightarrow W=mgs\\\Rightarrow W=mg15[/tex]
From Work Energy theorem
[tex]W=\frac{1}{2}m(v^2-u^2)\\\Rightarrow 2W-mv^2=-mu^2\\\Rightarrow mu^2=mv^2-2W\\\Rightarrow mu^2=mv^2-2(mgs)\\\Rightarrow u^2=v^2-2gs\\\Rightarrow u=\sqrt{v^2-2gs}\\\Rightarrow u=\sqrt{25^2-2\times -9.81\times 15}\\\Rightarrow u=30.32\ m/s[/tex]
Initial velocity of the rock is 30.32 m/s
[tex]mgs=\frac{1}{2}m(v^2-u^2)\\\Rightarrow gs=\frac{1}{2}(-u^2)\\\Rightarrow s=\frac{\frac{1}{2}(-u^2)}{g}\\\Rightarrow s=\frac{\frac{1}{2}(-30.32^2)}{-9.81}\\\Rightarrow s=46.855\ m[/tex]
Maximum height that the rock reached is 46.855 m
A. The rock’s speed just as it left the ground is 18.19 m/s
B. The maximum height reached by the rock is 16.88 m
Wd = mgh = ½m(v² – u²)
mgh = ½m(v² – u²)
Cancel m from both side
gh = ½(v² – u²)
Cross multiply
2gh = v² – u²
2 × 9.8 × 15 = 25² – u²
294 = 625 – u²
Collect like terms
u² = 625 – 294
u² = 331
Take the square root of both side
u = √331
u = 18.19 m/s
v² = u² – 2gh
v² – u² = –2gh
Divide both side by –2g
h = (v² – u²) / –2g
h = (0² – 18.19²) / (–2 × 9.8)
h = 16.88 m
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