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2 years ago

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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].

1 answer:

mario62 [17]2 years ago

5 0

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

$39\ ft/s=\frac{39}{3.281}=11.88\ m/s$

Equation of motion

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

Acceleration of the ball is 9.801 m/s²

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