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A 2 kg ball is dropped from a height of 60 m. Find its speed just before it hits the ground. Assume no friction. Answer​

Respuesta :

saidalhasni003

Answer:

-34.3m/s

Explanation:

first lets find the time befor it hit the ground by using free fall equation and we know we use that in one condition which is a constant acceleration in this case its a gravitational acceleration which is -9.8

[tex]h = \frac{1}{2} g {t}^{2} [/tex]

[tex]t = \sqrt{ \frac{2 \times 60}{9.8 } } = 3.5s[/tex]

now we know that the initial velocity its zero :

so applu the another kinematic equation which is

[tex]v = u - gt[/tex]

v is final velocity and u is the initial velocity and its equal zero.

v = - 9.8 × 3.5 = - 34.3

SoaringSunshine

[tex]\huge\sf\red{A}\pink{N}\orange{S}\green{W}\blue{E}\gray{R}[/tex]

Given,

mass of the ball, m=2 kg ;

height of the ball, h=60 m

The initial potential energy of the ball,

[tex]{\boxed{\underline{\sf{E_{p}=m g h=(2)(10)(60)=1200 \mathrm{~J}}}}[/tex]

When the ball reaches the ground, its potential energy becomes zero as it is entirely converted into its kinetic energy ( [tex]{\sf{E_{K}[/tex] ), i.e.,

[tex]{\Rightarrow{\sf{E}_{\sf{k}}=1200 \mathrm{~J}}}[/tex]

If v is the velocity attained by the ball just before reaching the ground,

[tex]{\Rightarrow{\sf{E}_{\sf{k}}=\frac{1}{2}{mv}^{2}}}}[/tex]

[tex]{\sf\Rightarrow{v=\sqrt{\frac{2 \sf{E}_{\sf{k}}}{\sf{m}}}=\sqrt{\frac{2 \times 1200}{2}}}[/tex]

[tex]{\Large{\mid{\underline{\overline{\Rightarrow{\sf{34.64\:m/s}}}}\:{\mid}[/tex]

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