Question

A stone is thrown vertically upwards with an initial velocity 20m/s. Find the maximum height it reaches and the time taken by it to reach the height.

(g= 10m/s)

who gives answer with explanation i will make him braillint

Answer 1

The maximum height it reaches will be 20 m.

The time taken by it to reach the height will be: t = 2 seconds.

Step-by-step explanation:

We know the equation of the motion under gravity

v² - u² = 2gs

u = initial velocity = 20 m/s

v = final velocity = 0 m/s

`s\:=\:h_(max)`

so

`v^(2) \:-\:u^(2) \:=\:2gs`

substituting u = 20 m/s, v = 0 m/s, g = -10 and
`s\:=\:h_(max)`

`\left(0\right)^2-\:\left(20\right)^2=\:2\left(-10\right)* h_(max)`

`2\left(-10\right)* h_(max)=-400`

`(2\left(-10\right)h_(max))/(-20)=(-400)/(-20)`

`\:h_(max)=20` m

Therefore, the maximum height it reaches will be 20 m.

We know the equation

u = gt

substitute u = 20, g = 10

20 = 10 × t

t = 20/10

t = 2 seconds

Therefore, the time taken by it to reach the height will be: t = 2 seconds.