Question

A sphere rolls off the top of stair way with a horizontal velocity of magnitude 200 cm/s. The steps are 10 cm high and 10 cm wide. Which step will the ball hit first. (g=10m/s²) *

Answer 1

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Ball rolls off the top of stairway. Horizontal velocity $\small \underline {\sf{u = 1 .8m/s  }}$

Since the stairs have the same width as height, lets find the place where horizontal and vertical distance become equal .

$\small \underline {\sf{ ut = gt^2/2; g = 10m/s^2}}$

$\small \underline {\sf{t = 2u / g = 0.36 }}$

Therefore horizontal distance covered

 $\small \underline {\sf{ = ut</p><p> = 0.648. }}$

Thus the ball will reach in $\small \underline {\sf{0.648/0.2 = 3.24 }}$ i.e the fourth step.

Answer 2

Answer:

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Answer:

Ball rolls off the top of stairway. Horizontal velocity \small \underline {\sf{u = 1 .8m/s }}

u=1.8m/s

Since the stairs have the same width as height, lets find the place where horizontal and vertical distance become equal .

\small \underline {\sf{ ut = gt^2/2; g = 10m/s^2}}

ut=gt

2

/2; g=10m/s

2

\small \underline {\sf{t = 2u / g = 0.36 }}

t=2u/g=0.36

Therefore horizontal distance covered

\small \underline {\sf{ = ut < /p > < p > = 0.648. }}

=ut</p><p>=0.648.

Thus the ball will reach in \small \underline {\sf{0.648/0.2 = 3.24 }}

0.648/0.2=3.24

i.e the fourth step.