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Solved Exa
Example 1. An iron ball of mass 3 kg is
released from a height of 125 m and falls
freely to the ground. Assuming that the
value of g is 10 m/s2, calculate
(i) time taken by the ball to reach the
ground
(ii) velocity of the ball on reaching the
ground
(iii) the height of the ball at half the time it
takes to reach the ground.​

Sagot :

pstnonsonjoku

According to the equations of motion, the time taken to reach the ground is 5 seconds.

Using;

s = ut + 1/2gt^2

s = distance

u = initial velocity

t = time taken

g = acceleration due to gravity

Note that u = 0 m/s since the object was dropped from a height

Substituting values;

125 = 1/2 × 10  × t^2

125 = 5t^2

t^2 = 125/5

t^2 = 25

t = 5 secs

Velocity on reaching the ground is obtained from

v = u + gt

Where u = 0 m/s

v = gt

v = 10 × 5

v = 50 m/s

At half the time it takes to reach the ground;

s =  ut + 1/2gt^2

Where u = 0 m/s

s =  1/2gt^2

s = 1/2  × 10  × (2.5)^2

s = 31.25 m

Learn more about equation of motion: https://brainly.com/question/8898885

AadhPrit

Answer:

(i) time taken by the ball to reach the

ground is 5 sec.

(ii) velocity of the ball on reaching the

ground is 50 m/s.

(iii) the height of the ball at half the time it

takes to reach the ground is 31.25 m.

Step-by-step explanation:

Solution :

(i) time taken by the ball to reach the

ground

[tex]\longrightarrow{\sf{ \: \: s= ut + \dfrac{1}{2} a{(t)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: 125= 0 \times t + \dfrac{1}{2} \times 10 \times {(t)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: 125= 0 + \dfrac{10}{2} \times {(t)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: 125= 0 + 5\times {(t)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: 125= 5\times {(t)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = \dfrac{125}{5}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = \dfrac{ \cancel{125}}{\cancel{5}}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = 25}}[/tex]

[tex]\longrightarrow{\sf{ \: \: t = \sqrt{25} }}[/tex]

[tex]\longrightarrow \: \: {\sf{\underline{\underline{\red{ t = 5 \: sec}}}}}[/tex]

Hence, the ball taken 5 sec to reach the ground.

[tex]\begin{gathered}\end{gathered}[/tex]

(ii) velocity of the ball on reaching the

ground

[tex]\longrightarrow{\sf{ \: \: {v}^{2} - {u}^{2} = 2as}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {v}^{2} - {0}^{2} = 2 \times 10 \times 125}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {v}^{2} = 20 \times 125}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {v}^{2} = 2500}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {v} = \sqrt{2500} }}[/tex]

[tex]\longrightarrow{\sf{ \: \: \underline{\underline{ \red{{v} = 50 \: m/s }}}}}[/tex]

Hence, the velocity of ball is 50 m/s.

[tex]\begin{gathered}\end{gathered}[/tex]

(iii) the height of the ball at half the time it

takes to reach the ground.

[tex]\longrightarrow{\sf{ \: \: s= ut + \dfrac{1}{2} a{(t)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= 0 \times \dfrac{5}{2} + \dfrac{1}{2} \times 10 \times { \left( \dfrac{5}{2} \right)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= 0 + \dfrac{10}{2} \times { \left( \dfrac{5}{2} \times \dfrac{5}{2} \right)}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times { \left( \dfrac{5 \times 5}{2 \times 2} \right)}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times { \left( \dfrac{25}{4} \right)}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times \dfrac{25}{4}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10 \times 25}{2 \times 4}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= \dfrac{250}{8}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= \dfrac{\cancel{250}}{\cancel{8}}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {\underline{\underline{\red{s= 31.25 \: m}}}}}}[/tex]

Hence, the height of the ball to reach the ground is 31.25 m.

[tex]\underline{\rule{220pt}{3.5pt}}[/tex]

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