Question

a ball throwen by a player reaches another player in 2s .wat is the maximum height attained by the ball above the point of projection take g as 10​

Answer 1

Answer

• Maximum height reached its 5 m

Explanation

Given

• Time = 2 sec
• G = 10 m/s²

To Find

• Maximum Height Reached

Solution

● T = (2u sinθ)/g

● g = 10 m/s²

✭ Value of u sinθ

→ T = (2u sinθ)/g

→ 2 = (2u sinθ)/10

→ 2 × 10 = 2u sinθ

→ 20 = 2u sinθ

→ 20/2 = u sinθ

→ u sinθ = 10 ❲ eq(1) ❳

✭ Maximum Height Reached

→ H = (u sinθ)²/2g

→ H = 10²/2×10

→ H = 100/20

→ H = 5 m

Answer 2

Answer:

$\huge \bf \: solution$

Time = 2 s = 2usin∅/g

G = 10 m/s

$\sf \: value \: of \: \sin \theta$

$\sf \: t \: = \dfrac{2u \sin \theta}{g}$

$\sf \: 2 = \dfrac{2u \sin \theta}{10}$

$\sf \: 2 \times 10 = 2u \sin \theta$

$\sf \: 20 = 2u \sin \theta \:$

$\sf \: u \sin \theta \: = \dfrac{20}{2}$

$\sf \: u \sin \theta = 10...... \: eq(1)$

Maximum height

$\sf h \: = \dfrac{2u \: { \sin}^{2} }{2g}$

$\sf \: \dfrac {10^2} {2 \times 10}$

$\sf \: \dfrac{100}{20}$

$\huge \bf height \small \: maximum \huge \: = 5 \: m$