Question

A ball is throwm up at speed is 4 m/s find the maximum highest reached by the ball​

Answer 1

GiveN :-

• A ball is throwm up at speed is 4 m/s.

To FinD :-

• The maximum highest reached by the ball.

SolutioN :-

Given that a ball is thrown with a speed of 4m/s .And we need to find the maximum highest reached by the ball . We know the formula to find the maximum height when initial velocity is given by ,

$\sf:\implies \pink{ Height_{(maximum)}=\dfrac{u^2}{2g} }\\\\\sf:\implies H_{(max)}= \dfrac{(4m/s)^2}{2(10m/s^2)}\\\\\sf:\implies H_{(max)}=\dfrac{16m^2/s^2}{20m/s^2}\\\\\sf:\implies H_{(max)}= \dfrac{4}{5} m \\\\\sf:\implies\underset{\blue{\tt Required\ Height}}{\underbrace{\boxed{\pink{\frak{ Height_{(maximum)}= 0.8 m }}}}}$

Answer 2

Here,

A ball is thrown up with a velocity of 4m/s, which is initial velocity (u) at maximum height point, final velocity (v) = 0 and when ball moves as thrown in upward direction then acceleration due to gravity acts on it in vertically downward direction which decreases the velocity of the ball, i.e, Retardation. Therefore, a = -9.

[Note: If upward direction is taken as +ve then downward direction is taken as -ve.]

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★ By using eqⁿ of motion,

• v² = u² + 2as

Where,

• v = 0
• u = 4m/s
• a = -9 ≃ 10m/s²
• s = Hₘₐₓ

Therefore,

⇒ (0)² = (4)² + 2 (-9) Hₘₐₓ

⇒ 0 = 16 -(2 × 10) Hₘₐₓ

⇒ 0 = 16 - 20 Hₘₐₓ

⇒ 20 Hₘₐₓ = 16

⇒ Hₘₐₓ = 16/20

⇒ Hₘₐₓ = 0.8m

• Hence, the maximum height reached by the ball is 0.8m.