Question

A ball is dropped from the the edge of a roof it takes 0.1s to cross a window of height 2.0m find the height of the roof above the top of the window

Answer 1

Answer:

The height of the roof above the top of the window is 19.4 m.

Explanation:

The ball is dropped from the edge of a roof. It takes 0.1 s to cross a window of height 2 m.

Time taken, t = 0.1 s

Height of the window, S = 2 m

Acceleration due to gravity on the ball, g = 9.8 m/s²

The velocity of the ball just above the window can be calculated using the 2nd equation of motion.

   S = ut + (gt²/2)

⇒ u = [S - (gt²/2)] / t

      = [2 - (9.8 × 0.1² /2)] / 0.1

      = [2 - 0.049] / 0.1

   u = 19.51 m/s

This velocity is the final velocity of the ball just above the window when it is dropped from the rooftop.

∴ Final velocity, v = 19.51 m/s

 Initial velocity, u = 0 m/s

Now from the 3rd equation of motion:

v² - u² = 2aS

S = (v² - u²) / 2a

  = (19.51² - 0²) / (2 × 9.8)

  = 380.64 / 19.6

S = 19.4 m