Question

the maximum height attained by a ball projected with speed 20m/s at an angle 45° with the horizontal is​

Answer 1

Answer:

• The maximum height (H) of the ball is 10 m.

Given:

1. Initial speed of the ball (u) = 20 m/s.
2. Angle of projection (θ) = 45°

Explanation:

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⇒ H = u² sin² θ / 2 g

⇒ H = (20)² × sin² (45°) / 2 × 10

⇒ H = 400 × (1 / √2)² / 20

⇒ H = 400 × 1 / 2 / 20

⇒ H = 400 / 20 × 2

⇒ H = 400 / 40

⇒ H = 10

⇒ H = 10 m.

∴ The maximum height (H) of the ball is 10 m.

Note:

• Symbols have their usual meaning.

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Extra Formulas:

⇒ R = u² sin 2θ / g

⇒ R tan θ = 4 H

⇒ H = g T² / 8

⇒ T = 2 u sinθ / g

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Answer 2

$\tt{\huge{\green{Solution_{KW} \:: - }}}$

Question :

The maximum height attained by a ball projected with speed 20m/s at an angle 45° with the horizontal is _____

Solution :

$\tt{\orange {Step-By-Step-Explaination \: :- }}$

Let H be the maximum height of the given ball.

Then H is equal to :

$\tt{ \purple{ \leadsto{H \: = \dfrac{ {u}^{2} { \sin( \phi) }^{2} }{2g} }}}$

Substituting the above values and solving, we get :

H = 10.2 Metres.

Hence we can conclude that the maximum height reached by the ball is 10.2 M.