Question

PLEASE HELP

Answer 1

The speed of the rock at 20 m is 34.3 m/s

Explanation:

We can solve this problem by using the law of conservation of energy: the mechanical energy of the rock, sum of its potential energy + its kinetic energy) must be conserved in absence of air resistance. So we can write:

$U_i +K_i = U_f + K_f$

where :

$U_i$ is the initial potential energy

$K_i$ is the initial kinetic energy

$U_f$ is the final potential energy

$K_f$ is the final kinetic energy

The equation can also be rewritten as  follows:

$mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where:

m = 100 kg is the mass of the rock

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 80$ is the initial height

u = 0 is the initial speed  (the rock starts at rest)

$h_f = 20 m$ is the final height of the rock

v is the final speed when h = 20 m

And solving for v, we find:

$v=\sqrt{2g(h_i-h_f)}=\sqrt{2(9.8)(80-20)}=34.3 m/s$

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