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.

[tex]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})} \\\\[/tex]

[tex]= 1.75 \ sec \\\\[/tex]

Distance:

[tex]x = v_{ix} \times t \\\\[/tex]

[tex]= 8 \times 1.75\\\\= 14 m[/tex]

fichoh fichoh · Answer 2 · 2021-11-12T07:13:20+01:00

Using the concept of motion, the answers to the question posed on the motion of a dropped ball are :

Vertical component of acceleration = - 9.8 m/s²
Horizontal component of acceleration = 0
Initial vertical Velocity = 0
Time taken = 1.75 seconds
Landing distance from building = 14 meters

The vertical component of acceleration is denoted as [tex]a_{y} [/tex] :

The gravitational force acting on the ball in the downward direction has an acceleration of - 9.8 m/s². Hence,

[tex]V_{y} = 9.8 m/s^{2} [/tex]

Gravitational force only acts downwards along the vertical, hence, horizontal component of acceleration will be Zero.

[tex]V_{y} = 0 [/tex]

then the vertical component of x, will be 0

[tex]V_{x} [/tex] = 0

The initial vertical velocity ; [tex]V_{iy} is Zero [/tex] :

[tex]V_{iy} = 0 [/tex] :

The time taken :

S = 0.5gt² ; S = distance = 15 meters ; g = - 9.81

15 = 0.5(9.8)t²

15 = 4.9t²

t² = (15 ÷ 4.9)

t² = 3.061

t = √3.061

t = 1.75 seconds

Landing Distance from building :

Horizontal speed × time taken

[tex]V_{ix} \times 1.75 [/tex]

[tex]8 \times 1.75 = 14 meters [/tex]

Therefore, the ball landed 14 meters away from the building.

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