Question

four
start?
5. A stone is thrown in a vertically
upward direction with a velocity
of 5 msl. If the acceleration of the
stone during its motion is 10 ms?
in the downward direction, what
will be the height attained by the
stone and how much time will it
take to reach there?
trad
hexa
in F
will
whil
trad
octa​

Answer 1

Answer:

Maximum height = 1.25 metres

Maximum height = 1.25 metresTime taken = 0.5 seconds

Step-by-step explanation:

Given:

Initial velocity = u = 5 m/s

Final velocity = v = 0 m/s ( As the stone would come to a stop at the end of the upward journey)

a = -10 m/s^2 ( As we are moving against the gravity)

To find :

Max height and time taken to reach max height

Using the third equation of motion, which says:

$v ^{2} - u^{2} = 2as$

Substituting the values,

${0}^{2} - {5}^{2} = 2 \times - 10 \times s$

$- 25 = 2 \times - 10 \times s$

$- 25 = - 20s$

$s = \frac{ - 25}{ - 20}$

$s = 1.25$

The maximum height obtained by the stone is 1.25 metres

Now we can find the time using the first equation of motion which says :

V=u+at

0=5-10t

-5= - 10t

$t = \frac{1}{2} = 0.5 \: seconds$

The time taken to reach 1.25 metres is 0.5 seconds

Answer 2

Answer:

⋆ DIAGRAM :

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(2,0){\vector(0,2){1.5}}\put(2,1.8){\circle*{2}}\put(2.3,1.3){\vector(0,-3){1}}\put(2.1,0.05){\sf{u = 5 m/s}}\put(2.1,1.4){\sf{v = 0 m/s}}\put(1.1,1.8){\bf{Stone}}\put(2.4,0.8){\sf{a = - 10 m/s^\text2 [Against Gravity]}}\end{picture}$

$\rule{130}{1}$

$\frak{Given}\begin{cases}\textsf{Initial Velocity (u) = 5 m/s}\\\textsf{Final Velocity (v) = 0 m/s}\\\textsf{Acceleration (a) = - 10 m/s ^\text2 }\end{cases}$

$\underline{\bigstar\:\textbf{Using Third Equation of Motion :}}$

$:\implies$ v² – u² = 2as

$:\implies$ (0)² – (5)² = 2 × – 10 × s

$:\implies$ 0 – 25 = – 20s

$:\implies$ – 25 = – 20s

• Dividing both term by – 20

$:\implies$ s = 1.25 m

⠀

$\therefore\:\underline{\textsf{Maximum Height attained will be \textbf{1.25 metre}}}.$

$\rule{170}{2}$

$\underline{\bigstar\:\textbf{Using First Equation of Motion :}}$

$\dashrightarrow\:$ v = u + at

$\dashrightarrow\:$ 0 = 5 + (– 10 × t)

$\dashrightarrow\:$ 0 = 5 – 10t

$\dashrightarrow\:$ 10t = 5

• Dividing both term by 5

$\dashrightarrow\:$ 2t = 1

$\dashrightarrow\:\sf t=\dfrac{1}{2}$

$\dashrightarrow\:$ t = 0.5 sec

⠀

$\therefore\:\underline{\textsf{Time taken will be \textbf{0.5 seconds}}}.$