Question

A particle is projected upwards with a velocity of 100 m/s at an angle 60 with the vertical. time taken by the particle to move perpendicular to its initial direction is​

Answer 1

ANSWER:


ANSWER
u
=100sin60
0

i
^
+100cos60
0

j
^
​
=50
j
^
​
+50
3
​

i
^

v
=
u
−gt
j
^
​
=50
j
^
​
+50
3
​

i
^
−10t
j
^
​
=50
3
​

i
^
+(50−10t)
j
^
​

When velocity is perpendicular to initial velocity,
v
.
u
=0, or (50
3
​

i
^
+(50−10t)
j
^
​
).(50
j
^
​
+50
3
​

i
^
)=0 or, 7500+2500−500t=0
which gives, t=20s

Hope it helps!!!

Answer 2

Explanation:

★ GIVEN QUESTION :-

A ball is projected upwards with a velocity of 100 m/sec ata an angle of 60° with the vertical. Find the time when the particle will move perpendicular to its initial direction, taking g = 10 m/sec².

______________________________________________

★ ANSWER :-

\begin{gathered}\begin{gathered}\\\;\sf{\;t = \bold{20 \: sec }}\end{gathered} \end{gathered}

t=20sec

→ A ball is projected upwards with a velocity of 100 m/sec at an angle of 60° with the vertical. It will take 20 sec when the particle will move perpendicular to its initial direction, taking g = 10 m/sec².

______________________________________________

★ SOLUTION :-

Given,

\begin{gathered}\begin{gathered}\\\;\sf{\odot\;\;velocity \: of \: ball\;=\;\bf{\green{100m/sec}}}\end{gathered} \end{gathered}

⊙velocityofball=100m/sec

\begin{gathered}\begin{gathered}\\\;\sf{\odot\;\; Angle \; formed=\;\bf{\green{60° \:with\: the \:vertical}}}\end{gathered} \end{gathered}

⊙Angleformed=60°withthevertical

To find ,

time when the particle will move perpendicular to its initial direction, taking g = 10 m/sec².

_____________________________________________

~ Let's solve it !!

\begin{gathered}\begin{gathered}\\\;\sf{:\Longrightarrow\;\;(ux(i) + uy(j)) \times (vx(i) + vy(j))\;=\;\bf{0}}\end{gathered} \end{gathered}

:⟹(ux(i)+uy(j))×(vx(i)+vy(j))=0

\begin{gathered}\begin{gathered}\\\;\sf{:\Longrightarrow\;\;(ux \times vx) + (uy \times vy)\;=\;\bf{0}}\end{gathered} \end{gathered}

:⟹(ux×vx)+(uy×vy)=0

we know that ,

uy = 100cos60° = 50m/s in which (gravity) g↓

ux = 100sin60° = 50√3 m/sec in which g = 0

v = constant × ux = vx

\begin{gathered}\begin{gathered}\\\;\sf{:\Longrightarrow\;\;ux ^{2} + 50 \: (50 - (g)10 \times t)\;=\;\bf{0}}\end{gathered} \end{gathered}

:⟹ux

2

+50(50−(g)10×t)=0

\begin{gathered}\begin{gathered}\\\;\sf{:\Longrightarrow\;\;50^{2} \times 3 + 2500 - 500t)\;=\;\bf{0}}\end{gathered} \end{gathered}

:⟹50

2

×3+2500−500t)=0

\begin{gathered}\begin{gathered}\\\;\sf{:\Longrightarrow\;\;2500( 3 + 1) =500t}\end{gathered} \end{gathered}

:⟹2500(3+1)=500t

\begin{gathered}\begin{gathered}\\\;\sf{:\Longrightarrow\;\;2500 \times 4 =500t}\end{gathered} \end{gathered}

:⟹2500×4=500t

\begin{gathered}\begin{gathered}\\\;\sf{:\Longrightarrow\;\;t = \dfrac{2500 \times 4}{500} }\end{gathered} \end{gathered}

:⟹t=

500

2500×4

\begin{gathered}\begin{gathered}\\\;\sf{:\Longrightarrow\;\;t = \bold{20 \: sec }}\end{gathered} \end{gathered}

:⟹t=20sec

\begin{gathered}\begin{gathered}\begin{gathered}\\\;\underline{\boxed{\tt{\odot\;\;Hence,\;\; T\;=\;\bf{\blue{20\; sec}}}}}\end{gathered} \end{gathered} \end{gathered}

⊙Hence,T=20sec