Physics, asked by mohitha93, 9 months ago

11. Suppose that the three balls shown in figure below start simultaneously from the tops.

of the hills. Which one reaches the bottom first ? explain​

Attachments:

Answers

Answered by ROHIT38sinhg
1

Answer:

1st

Explanation:

distance is lowest

Answered by 12752
2

Answer:1) path AFB : angles : tanΘ1=1/3,tanΘ1=3,sinΘ1=1/  

​  

10,sinΘ1=3/  

​  

10

v1 at F=gsinΘ1∗t1, v1^2=2gH/4=gH/2,v1=  

​  

(gH/2)

t1=  

​  

(H/2g)/(sinΘ1)     t1=  

​  

10∗sqrt(H/2g)

v2^2 = v1^2+2g3H/4 = 2gH  sqrt(2gH) =sqrt(gH/2)+gsinΘ2∗t2

t2=(  

​  

10/3)  

(H/2g)

​  

=(  

​  

10/3)  

(H/2g)

​  

 

t1+t2=2/3∗sqrt(10)∗sqrt(2H/g)=2.108∗  

​  

(2H/g)

2) path ADB:

angle tanΘ=1sinΘ=1/  

​  

2

v2=sqrt(2gH)=gsinΘ∗t⇒t=1.414  

​  

(2H/g)

3) path AEB: angle: tanΘ1=3,tanΘ2=1/3,sinΘ1=3/  

​  

10,tanΘ2=1/  

​  

10

v1=gsinΘ1∗t1=  

​  

(2g3H/4),t1=  

​  

10∗  

​  

(H/2g)/  

3

​  

 

v2=  

​  

(2gH)=  

​  

(3gH/2)+gsinΘ2∗t2,t2=  

​  

10(H/2g)[2−  

3

​  

]

t1+t2=  

​  

10∗  

​  

(2H/g)(1−1/  

3

​  

),t=1.33  

​  

(2H/g)

Explanation:So we can see that the ball along the path AFB takes longest time

to reach B. The ball along single straight line path, ADB takes less

time duration to reach B. For the path AEB, the time duration reduces

further to a minimum. But after a particular point, as the slope of AE

increase, the slope of EB reduces and the total time duration to  

reach B increase.

Similar questions