Question

A particle is thrown up vertically with a velocity of 50m/s.(a) What will be its velocity at the highest point of its journey?(b) How high the particle would rise?(c ) What time would it take to reach the highest point?
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Answer 1

Explanation:

u=50m/s

g=−10m/s

2

The velocity at the highest point will be zero

Thus, v=0

Height can be calculated as,

v

2

−u

2

=2gs

0−2500=−20s

s=125m

Time can be calculated as,

v=u+gt

0=50+(−10)t

t=5sec

Answer 2

$\huge \red {\tt{Answer:}}$

$\large \bold{Given:- }$

$\sf \rightarrow{Initial \: velocity(u) = 50m/s}$

$\sf \rightarrow{gravity(g) = - 10m/ {s}^{2} }$

$\large \bold{To \: find:}$

$\sf \leadsto{ Final\: velocity \: of \: the \: highest \: point \: of \: the \: journey.}$

$\sf \leadsto{Height \: of \: particle.}$

$\sf \leadsto{Time \: taken \: to \: reach \: the \: highest \: point.}$

$\large \bold{Concept \: used:}$

$\sf \longrightarrow{The \: velocity \: at \: the \: highest \: point \: will \: be \: zero.}$

$\large \bold{Formula \: used:}$

$\mapsto \tt{{ {v}^{2} − {u}^{2} =2gs}}$

$\mapsto \tt{v =u + gt}$

$\large \bold{According \: to \: question:}$

$\fbox{Final \: velocity \: of \: the \: particle \: is \: zero.}$

$\sf \longmapsto{Then,we \: calculate \: height \: of \:the \: particle.}$

$\sf\implies{ {v}^{2} - {u}^{2} = 2gs }$

$\sf \implies{ {0}^{2} - {50}^{2} = 2 \times - 10 \times s}$

$\sf \implies{0} - 2500 = - 20s$

$\sf \implies{ \frac{ - 2500}{ - 20} = s = 125m }$

$\fbox{Hence,the \: maximum \: height \: of \: the \: particle \: is \: 125m.}$

$\sf{⇒Time \: taken \: to \: reach \: maximum \: height \: is,}$

$\sf \implies{v = u + gt}$

$\sf \implies{0 = 50 + ( - 10)t}$

$\sf \implies{t = 5sec}$

$\fbox{Time \: taken \: to \: reach \: maximum \: height \: is \: 5sec.}$

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