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A ball is thrown horizontally from the top of a building 18.4 m high. The ball strikes the ground at a point 79.2 m from the base of the building. The acceleration of gravity is 9.8 m/s. Find the time the ball is in motion. Answer in units of s.

Respuesta :

Answer:

1.94 s

Explanation:

To find the time of flight of the ball, we can just analyze its vertical motion.

The motion of the ball is a free fall motion, which is a uniform accelerated motion. Therefore we can use the suvat equation:

[tex]s=ut+\frac{1}{2}at^2[/tex]

where

s is the vertical displacement

u = 0 is the initial vertical velocity of the ball

t is the time

[tex]a=g[/tex] is the acceleration of gravity

Chosing downward as positive direction,

s = 18.4 m

[tex]g=9.8 m/s^2[/tex]

And solving for t, we find the time during which the ball is in motion:

[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(18.4)}{9.8}}=1.94 s[/tex]