Question

Q1.Invitialy of projectile “ (30î+40î)m/sec
Then find it angle of projection, flight time,
height
and horizontal range (g=10 m/second ²​

Answer 1

Step-by-step explanation:

Given :

$\begin{gathered}\leadsto V = (30\hat{i} + 40 \hat{j}) m/s \\ \\\end{gathered} </p><p>$

Horizontal component of velocity = Vₓ = u cos Ф = 30 m/s

Vertical component of velocity = Vᵧ = u sin Ф = 40 m/s

Acceleration due to gravity = g = 10 m/s²

Solution :

(1) Angle of projection (Ф )

⇒ tan Ф = Vᵧ / Vₓ

⇒ tan Ф = 40/30

⇒ tan Ф = 4 / 3

⇒ Ф= 53°

(2) Time of flight (T)

⇒ T = 2 u sin Ф /g

⇒ T = 2 x Vᵧ / g

⇒ T = 2 x 40 / 10

⇒ T = 8 s

(3) Maximum height (H)

⇒ H = u² sin ² Ф / 2g

⇒ H = u sin Ф x u sin Ф / 2 g

⇒ H = Vᵧ x Vᵧ / 2 x g

⇒ H = 40 x 40 / 2 x 10

⇒ H = 40 x 2

⇒ H = 80 m

(4) Horizontal range (R)

⇒ R = u² sin 2Ф / g

we know that ,

sin 2Ф = sinФ. cosФ

Hence ,

⇒ R = u² sinФ . cos Ф / g

⇒ R = u sin Ф . u cos Ф / g

⇒ R = Vᵧ . Vₓ / g

⇒ R = 40 x 30 / 10

⇒ R = 120 m

Answers :

• The angle of projection is 53 °
• The time of flight of the projectile is8 s.
• The maximum height of the projectile is 80 m .
• The horizontal range of the projectile is 120 m .