Question

A stone is thrown in vertically upward direction with a velocity of 5m/s. If the acceleration of the stone during its motion is 10m/s2 in thedownward 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:

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

• Kindly go to web for the diagram.

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 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!

→ v = u + at

→ 0 = 5 + (-10)(t)

→ 0 = 5 + (-10t)

→ 0 - 5 = -10t

→ -5 = -10t

→ 5 = 10 t

→ 5/10 = t

→ 0.5 = t

→ t = 0.5 seconds

→ Time = 0.5 seconds

~ Now let's calculate height!

→ v² - u² = 2as

→ (0)² - (5)² = 2(-10)(s)

→ 0 - 25 = (-20s)

→ -25 = -20s

→ 25 = 20s

→ 25/20 = s

→ 5/4 = s

→ 1.25 = s

→ s = 1.25 metres

→ Distance = 1.25 metres