Physics, asked by Anonymous, 6 months ago

Pls solve this.....
Spamming not allowed ❌❌❌
spam answers will be reported immediately
pls solve with full explanation .​

Attachments:

Answers

Answered by Anonymous
18

Given:-

  • Final Velocity of ball = 0m/s

  • Acceleration due to gravity = -9.8m/s²

  • Time = 6s

To Find:-

  • The velocity with which it was thrown up.

  • The maximum height it reaches

  • Its Position after 4s

Formulae used:-

  • v = u + gt

  • v² - u² = 2gh

  • h = ut + ½ × g × t²

Where,

  • v = Final Velocity
  • u = Initial Velocity
  • g = Acceleration due to gravity
  • t = Time
  • h = Height

Now,

As it is given that it takes 6 seconds to returns to thrower. it must have taken 3 seconds to reach the maximum height.

Therefore,

→ v = u + gt

→ 0 = u + -9.8 × 3

→ -u = -29.4

→ u = 29.4m/s

Hence, The Initial velocity with which ut was thrown up is 29.4m/s

Now,

→ v² - u² = 2gh

→ (0)² - (29.4)² = 2 × -9.8 × h

→ -864.36 = -19.6h

→ h = -864.36/-19.6

→ h = 44.1m

Hence, The maximum height reached by ball is 44.1m

Now, after reaching maximum height it's initial Velocity becomes zero. at 3 second it was on height 44.1 which is it's maximum Height. after that the ball will start falling freely. Hence, we will have to find distance travelled in 1 second and then subtract it with maximum height

Therefore,

→ s = ut + ½ × a × t²

→ s = 0 × 1 + ½ × 9.8 × (1)²

→ s = 0 + 4.9 × 1

→ s = 4.9m

Therefore, The height travelled in that 1 second is 4.9m

Hence, it's Position after 4 second is 44.1 - 4.9 = 39.2m.

Answered by saritabakshig90
3

Answer:

ᵂᴸᶜᴹ..ˢⁱˢᵒ

ᵇᵉ ʰᵃᵖᵖʸ..ˣᵈ☺️

Similar questions