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Answers
◘ In the first case, (when going upward)
• Initial speed =
• Acceleration while ball is moving upward :-
We know,
◘ In the second case, (when coming down)
•
•
Now,
Putting the values of a and s :-
Hence, the answer is option (a).
Answer:
◘ In the first case, (when going upward)
• Initial speed = \sf{v_o}v
o
• Acceleration while ball is moving upward :-
\sf{\dfrac{w+f}{m}}
m
w+f
We know,
\sf{v_f^2=v_i^2 +2\overrightarrow{a}.\overrightarrow{s}}v
f
2
=v
i
2
+2
a
.
s
\begin{gathered}\sf{\implies 0=v_o^2-2s\bigg(\dfrac{w+f}{m} \bigg)}\\\\\sf{\implies -v_o^2=-2s\bigg(\dfrac{w+f}{m} \bigg)}\\\\\sf{\implies v_o^2=2s\bigg(\dfrac{w+f}{m} \bigg)}\\\\\sf{\implies 2s=\dfrac{v_o^2}{\bigg(\dfrac{w+f}{m} \bigg)} }\\\\\sf{\implies s=\dfrac{v_o^2}{2\bigg(\dfrac{w+f}{m} \bigg)} }\end{gathered}
⟹0=v
o
2
−2s(
m
w+f
)
⟹−v
o
2
=−2s(
m
w+f
)
⟹v
o
2
=2s(
m
w+f
)
⟹2s=
(
m
w+f
)
v
o
2
⟹s=
2(
m
w+f
)
v
o
2
◘ In the second case, (when coming down)
• \sf{v_i=0}v
i
=0
• \sf{a=w+f}a=w+f
Now,
\sf{v_f^2=v_i^2+2 \overrightarrow{a}.\overrightarrow{s}}v
f
2
=v
i
2
+2
a
.
s
Putting the values of a and s :-
\begin{gathered}\sf{v_f^2=\dfrac{2\times (w-f)v_o^2}{2(w+f)} }\\\\\sf{\hookrightarrow v_f^2=\dfrac{ (w-f)v_o^2}{(w+f)} }\\\\\sf{\hookrightarrow v_f= \sqrt{\dfrac{(w-f)v_o^2}{(w+f)} }}\\\\\sf{\hookrightarrow v_f= v_o\sqrt{\dfrac{(w-f)}{(w+f)} } }\\\\\boxed{\boxed{\sf{\color{magenta}{\hookrightarrow v_f= v_o\bigg(\dfrac{w-f}{w+f} \bigg)^\frac{1}{2}}}}}\end{gathered}
v
f
2
=
2(w+f)
2×(w−f)v
o
2
↪v
f
2
=
(w+f)
(w−f)v
o
2
↪v
f
=
(w+f)
(w−f)v
o
2
↪v
f
=v
o
(w+f)
(w−f)
↪v
f
=v
o
(
w+f
w−f
)
2
1
Hence, the answer is option (a).
Please brainlest me the answer