Question

A ball is thrown vertically upwards with velocity 20 m/s. find the time taken for the ball to reach a height of 20m

Answer 1

⇝ Given :-

• A ball is thrown vertically upwards with velocity 20 m/s.

• It reaches a height of 20 m.

⇝ To Find :-

• Time taken by the ball to reach a height of 20 m.

⇝ Solution :-

Here,

• Initial Velocity = u = 20 m/s

• Acceleration = a = - g = - 10 m/s²

• Height = s = 20 m

Let Time Taken by Ball = t second

❒ We Have 2nd Equation of Motion as :

$\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{s = ut + \frac{1}{2}at {}^{2} }}}$

✧ Putting Given Values In above ;

$:\longmapsto \rm 20 = 20 \times t + \frac{1}{2} \times ( - 10) \times t {}^{2}$

$:\longmapsto \rm 20 = 20t - 5t {}^{2}$

$:\longmapsto \rm 5t {}^{2} - 20t + 20 = 0$

$\maltese \: \frak{Dividing \: \text Whole \: Equation \: b\text y \: 5}$

$:\longmapsto \bf t {}^{2} - 4t + 4 = 0$

$:\longmapsto \rm t(t - 2) - 2(t - 2) = 0$

$:\longmapsto \rm (t - 2)(t - 2) = 0$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf t = 2 \: s} }}}$

Therefore,

$\large\underline{\pink{\underline{\frak{\pmb{\text Time \: \text Taken = 2 \: second}}}}}$

⇝ Additional Information :-

✿ Three Equations of Motion :

$\blue{\large \qquad \boxed{\boxed{\begin{array}{cc} \maltese \: \: \bf v = u + at \\ \\ \maltese \: \: \bf s = ut + \dfrac{1}{2}a {t}^{2} \\ \\ \maltese \: \: \bf{v}^{2} - {u}^{2} = 2as\end{array}}}}$

Answer 2

Given :-

A ball is thrown vertically upwards with velocity 20 m/s

To Find :-

Time taken for the ball to reach a height of 20m

Solution :-

We know that

s = ut + 1/2 at²

20 = 20 × t + 1/2 × g × t²

20 = 20t + 1/2 × -10 × t²

20 = 20t + (-5)t²

20 - 20t + 5t² = 0

Rearrange

5t² - 20t + 20

5(t² - 4t + 4) = 0

t² - 4t + 4 = 0

t² - (2t + 2t) + 4 = 0

t² - 2t - 2t + 4 = 0

t(t - 2) - 2(t - 2) = 0

(t - 2)(t - 2) = 0

t - 2 = 0

t = 0 + 2

t = 2 sec