Respuesta :

MrRoyal

Answer:

3.5 s

Explanation:

Given

[tex]Initial\ Velocity (u) = 0m/s[/tex] --- because it starts from rest

[tex]Height\ (S) = 60\ m[/tex]

Required

Determine time (t) to hit the ground

This can be solved using the following motion equation

[tex]S = ut + \frac{1}{2}gt^2[/tex]

Substitute values for S and u and take g as 9.8m/s^2

[tex]60 = 0 * t + \frac{1}{2} * 9.8 * t^2[/tex]

[tex]60 = 0 + \frac{1}{2} * 9.8 * t^2[/tex]

[tex]60 = \frac{1}{2} * 9.8 * t^2[/tex]

Multiply through by 2

[tex]2 * 60 = 2 * \frac{1}{2} * 9.8 * t^2[/tex]

[tex]120 = 9.8 * t^2[/tex]

Solve for t^2

[tex]t^2 = \frac{120}{9.8}[/tex]

[tex]t^2 = 12.2448979592[/tex]

Take square roots

[tex]t = \sqrt{12.2448979592[/tex]

[tex]t = 3.49927106112[/tex]

[tex]t=3.5\ s[/tex] ---- approximated

samueladesida43

It would take a stone released from a height of 60m, 3.5s to hit the ground.

EQUATION OF MOTION:

  • The time taken for a stone of height 60m to hit the ground can be calculated by using the following formula of motion:

  • S = ut + ½at²

Where;

  1. S = distance (m)
  2. u = initial velocity (m/s)
  3. t = time
  4. a = acceleration due to gravity (9.8m/s²)

  • Since the stone is released from rest, it would have a initial velocity of 0m/s.

  • 60 = (0×t) + ½ × 9.8 × t²

  • 60 = 4.9t²

  • t² = 60 ÷ 4.9

  • t² = 12.24

  • t = √12.24

  • t = 3.5s

  • Therefore, it would take a stone released from a height of 60m, 3.5s to hit the ground.

Learn more at: https://brainly.com/question/13883874?referrer=searchResults

Otras preguntas