Question

A student on roof 49m height drops a stone one second later he throws a second stone at first they both hit the ground at same time , find the speedd did he throw the second stone. pleaase answer quick it's too urgent❤

Answer 1

$\huge{\mathcal{\underline{\underline{Given:-}}}}$

• one Stone is dropped from height 49m From the same height another stone is thrown after 1 sec
• Both stones reach the ground at the same time

$\huge{\mathcal{\underline{\underline{To\:find:-}}}}$

• The velocity with which the second stone was thrown (initial velocity of second stone)

$\huge{\mathcal{\underline{\underline{Solution:-}}}}$

In case of First stone ,

u = 0 (Since it is dropped)

h = 49m

a = g = 9.8 m/s²

time = t

From second equation of motion,

$\large\rm\bold{\underline{\boxed{S\:=\:ut+\frac{1}{2}at^2}}}$

$\large\rm{s\rightarrow{height}}$

$\large\rm{g\rightarrow{acceleration\:due\:to\:gravity}}$

$\large\rm{u\rightarrow{Initial\:velocity}}$

$\large\rm{t\rightarrow{Time}}$

$\large\rm{\implies{49\:=\:0(t)+\frac{9.8\times\:t^2}{2}}}$

$\large\rm{\implies{49\:=\:0+4.9t^2}}$

$\large\rm{\implies{10\:=\:t^2}}$

$\large\rm{\implies{t\:=\:3.16\:sec}}$

In case of second stone ,

Initial velocity = u₂ (let)

Time = t - 1 = 3.16 - 1 = 2.16 sec

Height = 49m

a = g = 9.8 m/s²

From Second equation of motion,

$\large\rm{\implies{49\:=\:(2.16)(u_2)+\frac{9.8\times\:(2.16)^2}{2}}}$

$\large\rm{\implies{49\:=\:2.16u_2+22.86}}$

$\large\rm{\implies{26.14\:=\:2.16u_2}}$

$\large\rm{\implies{u_2\:=\:12.1\:m/s}}$

∴ The velocity with which the second stone was thrown is 12.1 m/s

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Answer 2

Explanation:

\huge{\mathcal{\underline{\underline{Given:-}}}}

Given:−

one Stone is dropped from height 49m From the same height another stone is thrown after 1 sec

Both stones reach the ground at the same time

\huge{\mathcal{\underline{\underline{To\:find:-}}}}

Tofind:−

The velocity with which the second stone was thrown (initial velocity of second stone)

\huge{\mathcal{\underline{\underline{Solution:-}}}}

Solution:−

In case of First stone ,

u = 0 (Since it is dropped)

h = 49m

a = g = 9.8 m/s²

time = t

From second equation of motion,

\large\rm\bold{\underline{\boxed{S\:=\:ut+\frac{1}{2}at^2}}}

S=ut+

2

1

at

2

\large\rm{s\rightarrow{height}}s→height

\large\rm{g\rightarrow{acceleration\:due\:to\:gravity}}g→accelerationduetogravity

\large\rm{u\rightarrow{Initial\:velocity}}u→Initialvelocity

\large\rm{t\rightarrow{Time}}t→Time

\large\rm{\implies{49\:=\:0(t)+\frac{9.8\times\:t^2}{2}}}⟹49=0(t)+

2

9.8×t

2

\large\rm{\implies{49\:=\:0+4.9t^2}}⟹49=0+4.9t

2

\large\rm{\implies{10\:=\:t^2}}⟹10=t

2

\large\rm{\implies{t\:=\:3.16\:sec}}⟹t=3.16sec

In case of second stone ,

Initial velocity = u₂ (let)

Time = t - 1 = 3.16 - 1 = 2.16 sec

Height = 49m

a = g = 9.8 m/s²

From Second equation of motion,

\large\rm{\implies{49\:=\:(2.16)(u_2)+\frac{9.8\times\:(2.16)^2}{2}}}⟹49=(2.16)(u

2

)+

2

9.8×(2.16)

2

\large\rm{\implies{49\:=\:2.16u_2+22.86}}⟹49=2.16u

2

+22.86

\large\rm{\implies{26.14\:=\:2.16u_2}}⟹26.14=2.16u

2

\large\rm{\implies{u_2\:=\:12.1\:m/s}}⟹u

2

=12.1m/s

∴ The velocity with which the second stone was thrown is 12.1 m/s

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