Answer:
The frog takes 8 jumps to reach top of well
Explanation:
Given data
Frog at bottom=17 foot
Each time frog leaps 3 feet
Frog has not reached the top of the well, then the frog slides back 1 foot
To Find
Total number of leaps the frog needed to escape from well
Solution
in 1 jump distance jumped=3+(-1)
=2 feet
=2×1 feet
The "-1" is because the frog goes back
Now After 2 jumps the distance jumped as:
Distance Jumped=2+2
Distance Jumped=2*2
=4 feet
Similarly after 7 jumps
Distance Jumped=2+2+......+2
Distance Jumped=2*7
=14 feet
Now after 8th jump the frog climbs but doesnot slide back as it is reached to the top of well.
So
Distance Jumped=(Distance Jumped after 7 jumps)+3
=14+3
=17 feet
The frog takes 8 jumps to reach top of well
The speed of the pin after the elastic collision is 9 m/s east.
<h3>
Final speed of the pin</h3>
The final speed of the pin is calculated by applying the principle of conservation of linear momentum as follows;
m1u1 + mu2 = m1v1 + m2v2
where;
- m is the mass of the objects
- u is the initial speed of the objects
- v is the final speed of the objects
4(1.4) + 0.4(0) = 4(0.5) + 0.4v2
5.6 = 2 + 0.4v2
5.6 - 2 = 0.4v2
3.6 = 0.4v2
v2 = 3.6/0.4
v2 = 9 m/s
Thus, The speed of the pin after the elastic collision is 9 m/s east.
Learn more about linear momentum here: brainly.com/question/7538238
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Hi there!
We can begin by deriving the equation for how long the ball takes to reach the bottom of the cliff.

There is NO initial vertical velocity, so:

Rearrange to solve for time:

Plug in the given height and acceleration due to gravity (g ≈ 9.8 m/s²)

Now, use the following for finding the HORIZONTAL distance using its horizontal velocity:
