Question

A stone is thrown in a vertically upward direction with a velocity of 10m/s. If the acceleration of the stone during its motion is 10m/s^2 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

Figure regards this question:

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

Provided that:

• Initial velocity = 10 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 = 5 m
• Time taken = 1 second

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!

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 0 = 10 + (-10)(t) \\ \\ :\implies \sf 0 = 10 + (-10t) \\ \\ :\implies \sf 0 = 10 - 10t \\ \\ :\implies \sf 0 - 10 \: = -10t \\ \\ :\implies \sf -10 \: = -10t \\ \\ :\implies \sf 10 \: = 10t \\ \\ :\implies \sf \dfrac{10}{10} \: = t \\ \\ :\implies \sf 1 \: = t \\ \\ :\implies \sf t \: = 1 \: second \\ \\ :\implies \sf Time \: taken \: = 1 \: second \\ \\ {\pmb{\sf{Therefore, \: solved!}}}$

~ Now let us calculate the height attained by using third equation of motion!

$:\implies \sf v^2 \: - u^2 \: = 2as \\ \\ :\implies \sf (0)^{2} - (10)^{2} = 2(-10)(s) \\ \\ :\implies \sf (0)^{2} - (10)^{2} = (-20)(s) \\ \\ :\implies \sf (0)^{2} - (10)^{2} = (-20s) \\ \\ :\implies \sf 0 - 100 = -20s \\ \\ :\implies \sf -100 = -20s \\ \\ :\implies \sf 100 = 20s \\ \\ :\implies \sf \dfrac{100}{20} \: = s \\ \\ :\implies \sf 5 \: = s \\ \\ :\implies \sf s \: = 5 \: m \\ \\ :\implies \sf Height \: = 5 \: metres \\ \\ {\pmb{\sf{Therefore, \: solved!}}}$

Answer 2

Answer:

I hope it's helpful to you

Explanation:

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