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a ball is thrown straight up from the ground with an unknown velocity. Its reaches its highest point after 3.5s. With what velocity did it leave the ground?



Answer :

MathPhys

Answer:

34.3 m/s

Explanation:

The ball is in free-fall, so it undergoes constant acceleration. The motion can be modeled using kinematic equations, also known as SUVAT equations. The equation we will use is:

v = u + at

where

  • u is the initial velocity
  • v is the final velocity
  • a is the acceleration
  • t is the time

At the maximum height, the final velocity is v = 0 m/s. The time is t = 3.5 s. Taking up to be positive, the acceleration is a = -9.8 m/s². Solving for the initial velocity:

v = u + at

0 m/s = u + (-9.8 m/s²) (3.5 s)

u = 34.3 m/s

The ball was thrown at 34.3 m/s.

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