Question

A ball is thrown vertically up words with an initial velocity of 19.8 m/s after what time will it return to the ground

Answer 1

Answer:

ATQ, height, s=19.6m

(i) Initial velocity, u=?

Now, when thrown vertically upwards,

Acceleration= −g=−9.8m/s^2

Also, at end point, final velocity= 0m/s

⇒v^2=u 2 +2as

So, (0)^2  =(u)^ 2+2(−9.8)(19.6)

⇒u=19.6m/s

(ii) For going upward,

v=u+at

⇒0=19.6+(−9.8)(t)

⇒t=2s

Total time= upward+ downward

⇒2+2=4s

(iii) For downward motion,

u=0m/s

v=?

a=+9.8m/s

t=2s

⇒v=u+at

=19.6m/s

Answer 2

$\sf{\large{\underline{\underline{EXPLANATION}}}}$

$\sf{Here}\begin{cases}\sf{u=19.8ms^{-1}}\\ \sf{s=0}\\ \sf{a=-g=-9.8ms^{-2}}\sf{t=??}\end{cases}$

$\textbf{\underline{Using equation of motion,,}}</p><p>$

$\mathtt{\boxed{\blue{s=ut+\frac{1}{2}at^{2}}}}$

$\sf{\implies{0=19.8t+\cancel\dfrac{-9.8}{2}t^{2}}}$

$\sf{\implies{0=19.8t-4.9t^{2}}}$

$\sf{\implies{0=\frac{-49t^{2}}{10}+\frac{198t}{10}}}$

$\sf{\implies{49t^{2}-198t=0}}$

$\sf{\implies{t(49t-198)=0}}$

$\sf{\implies{t=0\:or\:t=\frac{198}{49}}}$

$\sf{\implies{t=0\:or\:t=4.04s}}$

Therefore,,

Time taken by ball=4.04s

Hope it helps ❣️❣️❣️