High School

A 0.25 kg object is projected vertically into the air with a velocity of 45 m/s. How high above the ground is the object after 4.3 seconds?

A) 103 m
B) 0 m
C) 87 m
D) 121 m​

Answer :

Final answer:

The height of the object after 4.3 seconds is found by utilizing the formula for displacement in kinematics, giving a result of approximately 87m. Therefore, the correct answer is (C).

Explanation:

To determine the height of the object after 4.3 seconds, we must use the formula for displacement in kinematics:

h = ut - 0.5gt^2

Where:

  • h is the height,
  • u is the initial velocity,
  • t is the time, and
  • g is the acceleration due to gravity, which is approximately 9.81 m/s^2.

In this case:

  • u = 45 m/s,
  • t = 4.3 s,
  • g = 9.81 m/s^2.

Substituting these values into the formula, we get:

h = (45 m/s)(4.3 s) - 0.5(9.81 m/s^2)(4.3 s)^2.

When calculated, this gives a height of approximately 87 m. Therefore, the object is approximately 87 m high after 4.3 seconds, making the correct answer choice (C).

Learn more about Physics of Vertical Projectiles here:

https://brainly.com/question/35283202

#SPJ11