Question

body is projected up with an initial velocity of 25 mt at an angle 35° with the horizontal Calculate (a) maximum height (b) time taken to reach the maximum height (c) horizontal range ​

Answer 1

Answer:

Explanation:

Given :

• Initial velocity (u) = 25 m/s
• Angle (θ) = 35°
• Acceleration due to gravity (g) = 9.8 m/s^2

To Find :

• (a) Maximum height.
• (b) Time taken to reach the maximum height.
• (c) Horizontal range.

Formula to be used :

• Maximum height formula, ie,.. H = u^2Sin^2θ/2g
• Horizontal range formula, ie,, R = u^2Sin2θ/g
• Time of flight formula, ie,, T = 2usinθ/g

Solution :

★Maximum height,

H = u^2Sin^2θ/2g

{ °.° Sin35° = 0.57 }

⇒ H = 25^2 × (0.57)^2 /2 × 9.8

⇒ H = 625 × 0.3249/19.6

⇒ H = 203.0/19.6

⇒ H = 10.3571428 ≈ 11

⇒ H = 11 m

★Time taken to reach the maximum height,

T = 2usinθ/g

⇒ T = 2 × 25 × sin35°/9.8

⇒ T = 50 × 0.57 /9.8

⇒ T = 28.5/9.8

⇒ T = 2.9 ≈ 3

⇒ T = 4 s

★Horizontal range,

R = u^2Sin2θ/g

⇒ R = (25)^2 × Sin2(35)°/9.8

⇒ R = 625 × Sin70°/9.8

{ °.° Sin70° = 0.93 }

⇒ R = 625 × 0.93/9.8

⇒ R = 581.25/9.8

⇒ R = 59.31122449 ≈ 60

⇒ R = 60 m

Answer 2

Given :-

• Initial velocity = 25 m/s
• Angle of projection = 35°

To Find :-

• Maximum height
• Time to reach maximum height ( Time of Ascent )
• Horizontal Range

Solution :-

• u = 25 m/s
• θ = 35°
• g = 9.8 m/s²

a) Calculating Maximum height :-

➩ $\sf Maximum \:height = \dfrac{u^2sin^2\theta}{2g}$

Substituting the values :-

➩ $\sf H = \dfrac{25^2 \times sin^2 35}{2 \times 9.8}$

➩ $\sf H = \dfrac{625 \times (0.57)^2}{2 \times 9.8}$

➩ $\sf H = \dfrac{203.0625}{19.6}$

➩ $\sf H = 10.36$

Maximum height = 10.36 m

b) Calculating time to reach maximum height -

➩ $\sf t = \dfrac{\frac{2usin\theta}{g}}{2}$

➩ $\sf t = \dfrac{u sin\theta}{g}$

Substituting the values :-

➩ $\sf t = \dfrac{25 \times sin35}{g}$

➩ $\sf t = \dfrac{25 \times 0.57}{9.8}$

➩ $\sf t = \dfrac{14.25}{9.8}$

➩ $\sf t = 1.45$

Time taken to reach maximum height = 1.45 sec

c) Calculating Horizontal Range :-

➩ $\sf R = \dfrac{u^2sin2\theta}{g}$

➩ $\sf R = \dfrac{25 \times 25 \times sin70}{9.8}$

➩ $\sf R = \dfrac{625 \times 0.93}{9.8}$

➩ $\sf R = \dfrac{581.25}{9.8}$

➩ $\sf R = 59.3$

Horizontal Range = 59.3 m