Question

a skateboarder is moving 5.25 m/s when he starts to roll up a frictionless hill. how much higher is he when he comes to a stop PLEASE HELP

Answer 1

The height reached by the skateboarder is 1.41 m

Explanation:

We can solve this problem by using the law of conservation of energy. In fact, the mechanical energy of the skateboarder (which is the sum of his potential energy + his kinetic energy) must be conserved. So we can write:

$U_i +K_i = U_f + K_f$

where

$U_i$ is the initial potential energy, at the bottom

$K_i$ is the initial kinetic energy, at the bottom

$U_f$ is the final potential energy, at the top

$K_f$ is the final kinetic energy, at the top

The equation can be rewritten as

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

where:

m is the mass of the skateboarder

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

$h_i = 0$ is the initial height

u = 5.25 m/s is the initial speed

$h_f$ is the maximum height reached by the skateboarder

v = 0 is the final speed (zero at the maximum height)

Solving for $h_f$, we find:

$\frac{1}{2}mu^2 = mgh_f\\h_f = \frac{u^2}{2g}=\frac{(5.25)^2}{2(9.8)}=1.41 m$

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