Question

a pebble is thrown vertically thrown upward with a speed of 20 m/s .how high will it be after 2s ​

Answer 1

Answer:

20 m

Explanation:

Initial velocity , u=20 m/s

Since, the ball is thrown upward, so 2nd equation of motion will apply,

S=ut− 1/2gt^2⇒S=20×2− 1/2×10×2^2

=40−20

=20 m

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Answer 2

Given-

• Initial Speed = 20m/s
• Time taken= 2s

   

Find-

• How High it will (Distance)

 

Formula to be used-

• $\boxed{\rm{ s= ut+\dfrac{1}{2} at^2 }}$

Where,

s= Distance

u= Initial speed

a= Acceleration

t= Time

 

Solution-

Using 2nd equation of motion,

$\sf{ s= ut+\dfrac{1}{2} at^2 }$

Put all given values in the formula we get,

$\bf{\implies s= 20\times 2 +\dfrac{1}{2}\times (-9.8) \times 2^2 }$

here a= acceleration of gravity that is g

where g= 9.8m/s²

$\bf{ \implies s= 40+\dfrac{1}{2} \times (-9.8) \times 4}$

$\bf{ \implies s= 40 + \dfrac{1}{2} \times (-39.2)}$

$\bf{ \implies s= 40- 19.6}$

$\bf{\implies s= 20.4 m}$

$\bf{\underline{\pink{\therefore Pebble \ coveres \ 20.4m \ in \ 2s.}}}$