Question

You throw a 3.00 N rock vertically into the air from ground level. You observe that when it is 15.0 m above the ground, it is traveling at 25.0 m/s upward. Use the work–energy theorem to find (a) the rock’s speed just as it left the ground and (b) its maximum height.

Answer 1

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

$W=Fs\\\Rightarrow W=mgs\\\Rightarrow W=mg15$

From Work Energy theorem

$W=\frac{1}{2}m(v^2-u^2)\\\Rightarrow 2W-mv^2=-mu^2\\\Rightarrow mu^2=mv^2-2W\\\Rightarrow mu^2=mv^2-2(mgs)\\\Rightarrow u^2=v^2-2gs\\\Rightarrow u=\sqrt{v^2-2gs}\\\Rightarrow u=\sqrt{25^2-2\times -9.81\times 15}\\\Rightarrow u=30.32\ m/s$

Initial velocity of the rock is 30.32 m/s

$mgs=\frac{1}{2}m(v^2-u^2)\\\Rightarrow gs=\frac{1}{2}(-u^2)\\\Rightarrow s=\frac{\frac{1}{2}(-u^2)}{g}\\\Rightarrow s=\frac{\frac{1}{2}(-30.32^2)}{-9.81}\\\Rightarrow s=46.855\ m$

Maximum height that the rock reached is 46.855 m