Question

$\large {\underline {\underline {\sf Question :}}}$
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 1

EXPLANATION.

Ball is projected upwards with a velocity = 100 m/sec.

at an angle of = 60°.

To find the time when the particle will move perpendicular to its initial direction.

Take = g = 10m/sec².

Initial velocity = u = 100m/s.

⇒ ∅ = 60°.

⇒ g = 10m/s².

Acceleration at x-axis = 0.

Initial velocity at x-axis = u sin∅ = 100 sin(60°).

Initial velocity at x-axis = 100 X √3/2. = 50√3 m/s.

From newton first equation of kinematics,

⇒ v = u + at.

⇒ vₓ = uₓ - gt.

⇒ vₓ = uₓ - 0.

⇒ vₓ = uₓ.

⇒ vₓ = 50√3 m/s.

Initial velocity at y-axis = u cos∅.

initial velocity at y-axis = 100 X cos(60°) = 50m/s.

From newton first equation of kinematics,

⇒ vₐ = uₐ + at.

⇒ vₐ = 50√3 - gt.

Both are perpendicular then,

⇒ u₀ = uₓ(i) + uₐ(j).

⇒ u₀ = 50√3(i) + 50(j).

⇒ v₀ = 50√3(i) + ( 50 - gt )(j).

As we know that both are perpendicular,

u X V.

⇒ ( 50√3(i) + 50(j) ) X ( 50√3(i) + ( 50√3 - gt (j))).

⇒ 7500 + 2500 - 500t = 0.

⇒ t = 10000/500.

⇒ t = 20 seconds.

Answer 2

★ 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.

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★ ANSWER :-

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

→ 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.

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★ SOLUTION :-

Given,

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

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

To find ,

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

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~ Let's solve it !!

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

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

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

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

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

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

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

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

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

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