Question

Pls answer fast thank you

Answer 1

Horizontal angle = Ø = 30°

Horizontal distance = x = 20 m

Vertical distance = y = 5 m

Using trajectory equation of Projectile Motion.

g = Acceleration due to gravity.

u = Initial velocity.

y = x tanØ -(1/2 g/u² cos²Ø) x².

5 = 20 × 1/√3 - ( 1/2 × 10/u² × √3/2) 20²

5 = 20/√3 - (20/3u²) 20²

5 = 20/√3 - 20³/3u²

20³/3u² = 20/√3 - 5

20³/3u² = 20-5√3/√3

20³/3 × √3/20-5√3 = u²

u² = 20³/√3 × 20+5√3/20² - 75

u² = 20³/√3 × 5(4+√3)/325

u² = 20².4/√3 × 5(4+√3)/65

u² = 20².4/√3 × (4+√3)/13

u = √20².4/√3 × (4+√3)/13

u = 40√(4+√3)/13√3 ms-¹ Answer.....

Answer 2

✪ᴀɴsᴡᴇʀ✪

• A) $u = 40\sqrt{\frac{4+\sqrt{3}}{13\sqrt{3}}}$

★ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ★

☆ɢɪᴠᴇɴ☆

• A projectile thrown at θ = 30º lands on a building 5m high and 20m away from point of projection.

✯ᴛᴏ ғɪɴᴅ✯

• Initial Speed of Projectile.

᯽ɪᴍᴘᴏʀᴛᴀɴᴛ ᴇǫᴜᴀᴛɪᴏɴs᯽

Trajectory Equation of Projectile

$\red{\boxed{y = [tan(θ)]x - [\frac{g}{2u²cos²(θ)}]x²}}$

Equation of Motion in Vertical and Horizontal Direction

$\blue{\boxed{(x-x₀) = u_{x}t + \frac{1}{2}a_{x}t²}}$

$\green{\boxed{(y-y₀) = u_{y}t + \frac{1}{2}a_{y}t²}}$

☆ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴ☆

⍟ᴜsɪɴɢ ᴛʀᴀᴊᴇᴄᴛᴏʀʏ ᴇǫᴜᴀᴛɪᴏɴ⍟

Here

θ = 30º, g = 10 m/s²

x = 20 m, y = 5 m

Using trajectory equation

$5 = \frac{20}{\sqrt{3}} - [\frac{10}{2\times u²(\frac{\sqrt{3}}{2})²}]20²$

$⇒5 = \frac{20}{\sqrt{3}} - \frac{20³}{3u²}$

$⇒\frac{20³}{3u²} = \frac{20}{\sqrt{3}}-5$

$⇒\frac{20³}{3u²} = \frac{20-5\sqrt{3}}{\sqrt{3}}$

$⇒\frac{20³}{3}\times \frac{\sqrt{3}}{20-5\sqrt{3}} = u²$

$⇒\frac{20².20}{\sqrt{3}}\times \frac{20+5\sqrt{3}}{20²-5²\times 3} = u²$

$⇒\frac{20².20}{\sqrt{3}}\times \frac{20+5\sqrt{3}}{325} = u²$

$⇒\frac{20².4}{\sqrt{3}}\times \frac{5(4+\sqrt{3})}{65} = u²$

$⇒\frac{20².2²}{\sqrt{3}}\times \frac{(4+\sqrt{3})}{13} = u²$

$⇒u = 40\sqrt{\frac{4+\sqrt{3}}{13\sqrt{3}}}$

᯽ᴜsɪɴɢ ʙᴀsɪᴄ ᴇǫᴜᴀᴛɪᴏɴ ᴏғ ᴍᴏᴛɪᴏɴ᯽

Here

$θ = 30º,x₀ = 0, y₀ = 0$

$x = 20 m, y = 5 m$

$a_{x} = 0, a_{y} = -g = -10m/s²$

$u_{x} = ucos(30º) = \frac{\sqrt{3}u}{2}$

$u_{y} = usin(30º) = \frac{u}{2}$

Using these values in above equation of motion, we get

Horizontal Direction

$20 = \frac{\sqrt{3}u}{2}t$

$⇒t = \frac{40}{\sqrt{3}u}....(1)$

Vertical Direction

$5 = \frac{ut}{2} - 5t²$

Substituting 't' from (1) we get

$⇒5 = \frac{u}{2}.\frac{40}{\sqrt{3}u}- 5.\frac{40²}{3u²}$

$⇒5 = \frac{20}{\sqrt{3}}-\frac{5\times 40²}{3u²}$

$⇒5 = \frac{20}{\sqrt{3}} - \frac{20³}{3u²}$

$⇒\frac{20³}{3u²} = \frac{20}{\sqrt{3}}-5$

$⇒\frac{20³}{3u²} = \frac{20-5\sqrt{3}}{\sqrt{3}}$

$⇒\frac{20³}{3}\times \frac{\sqrt{3}}{20-5\sqrt{3}} = u²$

$⇒\frac{20².20}{\sqrt{3}}\times \frac{20+5\sqrt{3}}{20²-5²\times 3} = u²$

$⇒\frac{20².20}{\sqrt{3}}\times \frac{20+5\sqrt{3}}{325} = u²$

$⇒\frac{20².4}{\sqrt{3}}\times \frac{5(4+\sqrt{3})}{65} = u²$

$⇒\frac{20².2²}{\sqrt{3}}\times \frac{(4+\sqrt{3})}{13} = u²$

$⇒u = 40\sqrt{\frac{4+\sqrt{3}}{13\sqrt{3}}}$

Thus, initial velocity of projectile is $u = 40\sqrt{\frac{4+\sqrt{3}}{13\sqrt{3}}}$

I hope you understood. I put a lot of effort to type all these complicated equation. Please give thanks and mark as brainliest answer.