Question

A physics student throws a ball thrown horizontally from the top of a building with a speed of 8 m/s. The student measures the height of the building to be 15m.

Answer 1

Answer:

Following are the solution to the given question:

Explanation:

Please find the complete question in the attached file.

$a_x = 0\\\\a_y = -g = -9.81\ \ \frac{m}{s^2} \\\\v_{iy} = 0 \\\\y = v_{iy} \times t+ \frac{1}{2ay} \times t^2 \\\\ 15 = \frac{1}{2} \times 9.8 \times t^2 \\\\t = \sqrt{(2 \times \frac{15}{9.8})}$

  $= 1.75 \ sec$

Distance:

$x = v_{ix} \times t$

  $= 8 \times 1.75\\\\= 14 m$