Question

A ball is thrown vertically upward
at 20 m/s. Calculate (a) how high
it goes (b) the time to vem to
return to its starting poient (g = 10 m/s 2)​

Answer 1

$\huge\sf\pink{Answer}$

☞ Maximum height is 20 m

☞ Total time to come back is 20 sec

━━━━━━━━━━━━━

$\huge\sf\blue{Given}$

✭ Initial Velocity (u) = 20 m/s

✭ Final Velocity (v) = 0 m/s

✭ Gravitational Acceleration (a) = -10 m/s²

━━━━━━━━━━━━━

$\huge\sf\gray{To \:Find}$

◈ Highest point reached?

◈ Time taken for the ball to return to the starting point?

━━━━━━━━━━━━━

$\huge\sf\purple{Steps}$

So here to find the highest point reached we shall use the third equation of motion, that is,

$\underline{\boxed{\sf v^2-u^2 = 2as}}$

Substituting the given values,

➝ $\sf 0^2-20^2 = 2\times -10 \times s$

➝ $\sf -400 = -20s$

➝ $\sf \dfrac{-400}{-20} = s$

➝ $\sf \red{s = 20 \ m}$

Then the time taken for it to come back to its initial position can be found with the help of the first equation of motion, that is,

$\underline{\boxed{\sf v=u+at}}$

Substituting the values,

➳ $\sf 0 = 20+(-10)(t)$

➳ $\sf -20 = -10t$

➳ $\sf \dfrac{-20}{-10} = t$

➳ $\sf \green{t=2\ sec}$

So here it would go once from down to its maximum height and then return back,so

»» $\sf Total \ time = 2+2$

»» $\sf \orange{Time = 4 \ sec}$

━━━━━━━━━━━━━━━━━━