Question

A particle is projected with velocity 20 m/s at angle of 45 degrees with horizontal find out maximum height

Answer 1

Required Answer:

Given:

A particle is projected with velocity 20 m/s at angle of 45 degrees with horizontal

Find:

Find out the maximum height.

Given:

→ u = 20 m/s

→ θ = 45°

→ h = u² sin² θ / 2g

→ ( 20 )² sin² 45° / 2 × 9.8

→ 100 / 9.8

→ 10.204 m

Therefore, 10.204 m is the maximum height of the ball.

Answer 2

$\Huge{\underline{\underline{\red{\mathfrak{Question :}}}}}$

A particle is projected with velocity 20 m/s at angle of 45 degrees with horizontal find out maximum height.

$\rule{200}{2}$

$\Huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

Given :

Velocity (v) = 20 m/s

Gravitational Force (g) = 10 m/s²

Angle (θ) = 45°

For maximum height we have formula :

$\LARGE{\boxed{\boxed{\red{\sf{h \: = \: \frac{v^2 \: Sin^2 \theta}{2g}}}}}}$

(Putting Values)

$\Large \leadsto {\sf{h \: = \: \frac{(20)^2 \: Sin^2 (45^{\circ})}{2(10)}}}$

$\Large \leadsto {\sf{h \: = \: \frac{400 \: \times \: \frac{1}{(\sqrt{2})^2}}{20}}}$

$\Large \leadsto{\sf{h \: = \: \frac{\cancel{400} \: \times \: \frac{1}{\cancel{2}}}{20}}}$

$\Large \leadsto {\sf{h \: = \: \frac{\cancel{200}}{\cancel{20}}}}$

$\Large \leadsto {\sf{h \: = \: 10}}$

$\Large \hookrightarrow {\boxed{\red{\sf{h \: = \: 10 \: m}}}}$