Question

A stone thrown up is caught by the thrower after 6s. How
high did it go and what was its position 4s after start?
(g=9.8m/s 2 ).

Answer 1

Answer:

Gɪᴠᴇɴ :

▪ Total time of flight = 6s

▪ Acc. due to gravity = 9.8m/s²

Tᴏ Fɪɴᴅ :

▪ Maximum height reached by ball.

▪ Position of ball after 4s.

Fᴏʀᴍᴜʟᴀ :

⇒ For a body thrown vertically up with initial velocity u,

፨ Total time of flight (T) :

$\bigstar\:\underline{\boxed{\bf{\red{T=\dfrac{2u}{g}}}}}$

፨ Maximum height reached (H) :

$\bigstar\:\underline{\boxed{\bf{\blue{H=\dfrac{u^2}{2g}}}}}$

Cᴀʟᴄᴜʟᴀᴛɪᴏɴ :

» Maximum height (H) :

$\Rightarrow\sf\:H=\dfrac{u^2}{2g}\\ \\ \Rightarrow\sf\:H=\dfrac{g^2T^2}{4\times 2g}\\ \\ \Rightarrow\sf\:H=\dfrac{gT^2}{8}\\ \\ \Rightarrow\sf\:H=\dfrac{9.8×36}{8}\\ \\ \Rightarrow\underline{\boxed{\bf{\green{H=44.1\:m}}}}\:\gray{\bigstar}$

» Position of ball at t = 4s :

$\mapsto\sf\:Time\:of\:ascent=\dfrac{Total\:time\:of\:flight}{2}\\ \\ \mapsto\sf\:Time\:of\:ascent=\dfrac{6}{2}=3s\\ \\ \dag\sf\:Ball\:will\:reach\:at\:maximum\:height\:at\:t=3s\\ \\ \mapsto\sf\:d=ut+\dfrac{1}{2}gt^2\\ \\ \mapsto\sf\:d=0+\dfrac{1}{2}g(4-3)^2\\ \\ \mapsto\sf\:d=0.5\times 9.8\times 1\\ \\ \mapsto\bf\:d=4.9m$

※ At t = 4s :

Position of ball from ground = 44.1-4.9 = 39.2m

Answer 2

Answer :

$\bold{Total\;time\;of\;flight\;,}\\\\\bold{T=\frac{2u}{g} ...(1)}\\\\\bold{Maximum\;height\;reached\;,}\\\\\bold{H=\frac{u^2}{2g} }\\\\\implies \bold{H=(\frac{gT}{2} )^2*\frac{1}{2g} }\\\\\implies \bold{H=\frac{gT^2}{8} }\\\\\implies \bold{H=\frac{9.8*36}{8} }\\\\\implies \bold{H=44.1\;m}$

$\bold{Time\;of\;ascent=\frac{Total\;Time}{2} }\\\\\implies\bold{Time\;of\;ascent=6/2=3sec}$

So the ball will reach maximum height at t = 3 seconds

$\implies \bold{s=ut+\frac{1}{2}gt^2 }\\\\\implies \bold{s=0+\frac{1}{2}*9.8*(4-3)^2}\\\\\implies \bold{s=0.5*9.8*1}\\\\\implies \bold{s=4.9\;m}$

So at Time , t = 4 sec position of ball from air is 4.9 m

Position of ball from ground = 44.1 - 4.9 m

⇒ 39.2 m