Physics: Se lanza una piedra de 3.00 N verticalmente hacia arriba desde el suelo. Se observa que, cuando está 15.0 m sobre el suelo, viaja a 25.0 m/s hacia arriba. Use el teorema trabajo-energía para determinar a) su rapidez en el momento de ser lanzada. b) su altura máxima

Se lanza una piedra de 3.00 N verticalmente hacia

jtamika31

03.02.2022

(a). The speed at the moment of being thrown is 30.41 m/s.

(b). The maximum height is 47.18 m.

Explanation:

Given that,

Weight of stone = 3.00 N

Height = 15 m

Speed = 25.0 m/s

(a). We need to calculate the speed at the moment of being thrown

Using work energy theorem

$W=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

$-mg\times d=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

Put the value into the formula

$-9.8\times15=\dfrac{1}{2}\times(v_{2}^2-v_{1}^2)$

$-2\times9.8\times15=25^2-v_{1}^2$

$-v_{1}^2=-300-25^2$

$v_{1}=\sqrt{925}$

$v_{1}=30.41\ m/s$

(b). We need to calculate the maximum height

Using work energy theorem

$[tex]W=\dfrac{1}{2}mv_{2}^2-\dfrac{1}{2}mv_{1}^2$

$mg\times d=\dfrac{1}{2}mv_{2}^2-\dfrac{1}{2}mv_{1}^2$

Here, $\dfrac{1}{2}mv_{2}^2$ =0

$-(mg)\times d=\dfrac{1}{2}mv_{1}^2$

$d=\dfrac{v_{1}^2}{2g}$

Put the value into the formula

$d=\dfrac{30.41^2}{2\times9.8}$

$d=47.18\ m$

Hence, (a). The speed at the moment of being thrown is 30.41 m/s.

(b). The maximum height is 47.18 m.

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