Question

g a stone with mass m=1.60 kg IS thrown vertically upward into the air with an initial kinetic energy of 470 J. the drag force acting on the stone throughout its flight is constant, independent of the velocity of the stone, and has a magnitude of 0.900 N. what is the maximum height reached by the stone?

Answer 1

Answer:

Height reached will be 28.35 m

Explanation:

Here we can use the work energy theorem to find the maximum height

As we know by work energy theorem

Work done by gravity + work done by friction = change in kinetic energy

$-mgh - F_f h = 0 - \frac{1}{2}mv_i^2$

now we will have

$-1.60(9.8)(h) - 0.900(h) = - 470$

$-16.58 h = -470$

$h = 28.35 m$

so here the height raised by the stone will be 28.35 m from the ground after projection in upward direction