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Answers
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.
Answer: