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A ball thrown straight up climbs for 3.0 sec before falling. Neglecting air resistance, with what velocity was the ball thrown?

Respuesta :

aachen

Answer:

Speed, u = 29.4 m/s

Explanation:

Given that, A ball thrown straight up climbs for 3.0 sec before falling, t = 3 s

Let u is speed with which the ball is thrown up. When the ball falls, v = 0

Using first equation of motion as :

v = u + at

Here, a = -g

So, u = g × t

[tex]u=9.8\times 3[/tex]

u = 29.4 m/s

So, the speed with which the ball was thrown is 29.4 m/s. Hence, this is the required solution.

asuwafohjames

The velocity at which the ball was thrown is 29.4 m/s.

To calculate the velocity at which the ball was thrown, we use the formula below.

Formula:

  • v = u+gt.............. Equation 1

Where:

  • v = Final  velocity of the ball
  • u = Initial velocity of the ball
  • g = acceleration due to gravity of the ball
  • t = time.

From the question,

Given:

  • v = 0 m/s (At maximum height)
  • g = -9.8 m/s
  • t = 3.0 s

Substitute these values into equation 1

  • 0 = u+3(-9.8)
  • 0 = u-29.4
  • u = 29.4 m/s

Hence, The velocity at which the ball was thrown is 29.4 m/s.

Learn more bout velocity here: https://brainly.com/question/25749514