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A ball is dropped and falls with an acceleration of 9.8 m/s2 downward. It hits the ground with a velocity of 49 m/s downward. How long did it take the ball to fall to the ground?

Respuesta :

VenusRetrograde
First write down what we know.

V=49m/s
Vo=initial velocity=0m/s
A=-9.8m/s²
t=?

Then find an equation that suits the data, in this case its v=vo+a*t

V=Vo+at
plug in the data
49=0+-9.8t
49=-9.8t
divide both sides by -9.8
-5=t

Answer=5 seconds
Lanuel

It would take the ball 5 seconds to fall to the ground.

Given the following data:

  • Acceleration = 9.8 [tex]m/s^2[/tex]
  • Final velocity = 49 m/s.
  • Initial velocity = 0 m/s (since the ball is starting from rest).

To find how long (time) it took the ball to fall to the ground, we would use the first equation of motion;

Mathematically, the first equation of motion is calculated by using the formula;

[tex]V = U + at[/tex]

Where:

  • A is the acceleration.
  • V is the final velocity.
  • U is the initial velocity.
  • t is the time measured in seconds.

Substituting the given parameters into the formula, we have;

[tex]49 = 0 + 9.8t\\\\49 = 9.8t\\\\t = \frac{49}{9.8}[/tex]

Time, t = 5 seconds.

Therefore, it would take the ball 5 seconds to fall to the ground.

Read more: https://brainly.com/question/8898885

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