A student drops a ball from the top of a tall building; it takes 2.9 s for the ball to reach the ground.
(a) What was the balls speed just before hitting the ground?
In m/s
(b) What is the height of the building? In m

Respuesta :

We have,

  • Initial velocity (u) = 0 m/s
  • Time taken (t) = 2.9s
  • Acceleration due to gravity (g) = + 10 m/s² [Down]

To calculate,

  • Final velocity (v)
  • Height (h)

Solution,

v = u + gt

→ v = 0 + 10(2.9)

v = 29 m/s [tex]\qquad[/tex] ( Ans )

And,

h = ut + ½gt²

→ h = 0(2.9) + ½ × 10 × (2.9)²

→ h = 5 × 8.41

h = 42.05 m [tex]\qquad[/tex] … ( Ans )

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