Question

5. A stone is thrown in a vertically
upward direction with a velocity
of 5 m sł. If the acceleration of the
stone during its motion is 10 ms2
in the downward direction what
will be the height attained by the
stone and how much time will it
take to reach there?​

Answer 1

Answer:

Here's your answer

Explanation:

Given => During upward motion

               Final velocity (v1) = 0m/s

               Initial velocity (u2) = 5m/s

               acceleration (a) = -10

               Time(t) = ?

During downward motion

Initial velocity (u2=v1) = 0m/s

acceleration (a) = 10m/s

Time (t)=?

displacement (S) = 0 (stone has come back to its original position)

According to question

for upward motion

we know that

v1=u1+at

0 = 5 + (-10)t

0 = 5 - 10t

10t = 5

t = 5/10

t = 1/2sec ----------------- (1)

for downward motion

S=ut+1/2at^2

0=1/2*10*t^2

0= 5t^2

t^2 = 0

t = 0 --------------------- (2)

total tie = (1)+(2)

1/2+0

1/2 sec

Answer 2

Figure regards this question:

$\setlength{\unitlength}{1mm}\begin{picture}(7,2)\thicklines\multiput(7,2)(1,0){55}{\line(3,4){2}}\multiput(35,7)(0,4){12}{\line(0,1){0.5}}\put(10.5,6){\line(3,0){50}}\put(35,60){\circle*{12}}\put(37,7){\large\sf{u = 5 m/s}}\put(37,55){\large\sf{v = 0 m/s}}\put(21,61){\large\textsf{\textbf{Stone}}}\put(43,40){\line(0, - 4){28}}\put(43,35){\vector(0,4){18}} {\pmb{\sf{BrainlyButterfliee}}}\put(24, - 3){\large\sf{ \sf g = -10 m/s^2 }}\put(48,30){\large\sf{H = ?}} \quad \put(18,30){\large\sf{t = ?}}\end{picture}$

Provided that:

• Initial velocity = 5 m/s
• Final velocity = 0 m/s
• Acceleration = -10 m/s²

Don't be confused!

⋆ Final velocity cames as zero because when it is thrown upwards then it will be stopped at the highest point.

⋆ We write acceleration in negative except of positive because the object is thrown in upward direction.

To calculate:

• Height attained by stone
• Time taken

Solution:

• Height attain by stone = 1.25 m
• Time taken = 0.5 seconds

Using concepts:

• Newton's first law of motion
• Newton's third law of motion

Using formulas:

• Newton's first law of motion is given by the mentioned formula:

• ${\small{\underline{\boxed{\sf{v \: = u \: + at}}}}}$

• Newton's third law of motion is given by the mentioned formula:

• ${\small{\underline{\boxed{\sf{2as \: = v^2 \: - u^2}}}}}$

Where, v denotes final velocity, u denotes initial velocity, a denotes acceleration, t denotes time taken, s denotes displacement or distance or height.

Required solution:

~ Firstly let us find out the time taken!

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 0 = 5 + (-10)(t) \\ \\ :\implies \sf 0 = 5 + (-10t) \\ \\ :\implies \sf 0 = 5 - 10t \\ \\ :\implies \sf 0 - 5 \: = -10t \\ \\ :\implies \sf -5 \: = -10t \\ \\ :\implies \sf 5 \: = 10t \\ \\ :\implies \sf \dfrac{5}{10} \: = t \\ \\ :\implies \sf 0.5 \: = t \\ \\ :\implies \sf Time \: taken \: = 0.5 \: seconds \\ \\ {\pmb{\sf{Therefore, \: solved!}}}$

~ Now let's find out the distance!

$:\implies \sf 2as \: = v^2 \: - u^2 \\ \\ :\implies \sf 2(-10)(s) \: = (0)^{2} - (5)^{2} \\ \\ :\implies \sf 2(-10)(s) \: = 0 - 25 \\ \\ :\implies \sf 2(-10s) \: = 0 - 25 \\ \\ :\implies \sf -20s \: = 0 - 25 \\ \\ :\implies \sf -20s \: = -25 \\ \\ :\implies \sf 20s \: = 25 \\ \\ :\implies \sf s \: = \dfrac{25}{20} \\ \\ :\implies \sf s \: = 1.25 \: m \\ \\ :\implies \sf Height \: = 1.25 \: metres \\ \\ {\pmb{\sf{Therefore, \: solved!}}}$