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An object is launched upwards at an initial speed of 5 m/s. What is the maximum height reached by the object from where it was launched? Enter the value using up to two decimal places without the units. Hint: Calculate how long in seconds it takes for the object to reach its maximum height, then use that calculated time for the free-fall



Answer :

MathPhys

Answer:

1.28 m

Explanation:

The object 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:

v² = u² + 2as

where

  • s is the displacement
  • u is the initial velocity
  • v is the final velocity
  • a is the acceleration

The initial velocity is u = 5 m/s. At the maximum height, the final velocity is v = 0 m/s. The acceleration is that due to gravity, a = -9.8 m/s².

v² = u² + 2as

(0 m/s)² = (5 m/s)² + 2 (-9.8 m/s²) s

s = 1.28 m

The maximum height of the object is 1.28 meters.

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