The best leaper in the animal kingdom is the puma, which can jump to a height of 3.7 m when leaving the ground at an angle of 45°. With what speed must the animal leave the ground to reach that height?

Explanation:

Let the velocity of puma be a.

Consider the vertical motion of puma

We have equation of motion v² = u² + 2as

Initial velocity, u = asin45

Acceleration, a = -9.81 m/s²  

Final velocity, v = 0 m/s  

Displacement, s = 3.7 m

Substituting  

v² = u² + 2as

0² = (asin45)² + 2 x -9.81 x 3.7

a² = 145.19

a = 12.05 m/s

Speed required is 12.05 m/s

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)