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
Answers
Answer:
1st
Explanation:
distance is lowest
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.