Question

A stone is thrown in a vertically upward direction with a velocity of 5 ms. 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?​

Answer 1

Explanation:

Given,

initial velocity = 5m/s

acceleration in the downward direction = -10m/s² (gravity)

Final velocity = 0m/s (as the stone will ultimately stop due to downward acceleration)

Using formula,

v = u + at

0 = 5 - 10t

-5 = -10t

t = 1/2

t = 0.5s

Using formula,

S = ut + 1/2at²

S = 5×0.5 - 1/2 × 10 × 0.5 × 0.5

S = 2.5 - 1.25

S = 1.25m

Maximum height reached = 1.25m

Time taken = 0.5s

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Answer 2

Maximum Height attained will be 1.25 metre.

Time taken will be 0.5 seconds.

Explanation :

Initial velocity - 5 m/s

Final velocity - 0 m/s

Acceleration = - 10 m/s²

To find :

Max height and time taken to reach max height.

Solution :

We know that :

$\bigstar\;\boxed{\bf v^{2} - u^{2} = 2as}}$

Therefore,

$\implies \sf {0}^{2} - {5}^{2} = 2 \times - 10 \times s\\ \\ \\ \implies \sf - 25 = 2 \times - 10 \times s\\ \\ \\ \implies \sf- 25 = - 20s\\ \\ \\\implies \sf s = \dfrac{ - 25}{ - 20}\\ \\ \\ \implies \bf s = 1.25$

∴ The maximum height obtained by the stone is 1.25 metres.

$\rule{300}{1.5}$

The time taken :

We know that :

$\bigstar\;\boxed{\bf V=u+at}}$

So,

$\implies \sf 0=5-10t\\ \\ \\\implies \sf -5= - 10t\\ \\ \\\implies \sf \dfrac{1}{2}\\ \\ \\\implies \bf 0.5 \: seconds.$

∴The time taken to reach 1.25 metres is 0.5 seconds.

$\rule{300}{1.5}$

• Diagram :

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