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If a ball is dropped from a height​ (H) its velocity will increase until it hits the ground​ (assuming that aerodynamic drag due to the air is​ negligible). During its​ fall, its initial potential energy is converted into kinetic energy. If the ball is dropped from a height of 720720 centimeters​ [cm], and the impact velocity is 3939 feet per second​ [ft/s], determine the value of gravity in units of meters per second squared ​[m/s2​].

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Answer:

9.801 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 39 ft/s

s = Displacement = 720 cm = 7.2 m

a = Acceleration

Converting to m/s

[tex]39\ ft/s=\frac{39}{3.281}=11.88\ m/s[/tex]

Equation of motion

[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{11.88^2-0^2}{2\times 7.2}\\\Rightarrow a=9.801\ m/s^2[/tex]

Acceleration of the ball is 9.801 m/s²

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