A projectile is launched vertically into the air with a velocity of 19.0 m/s. Calculate the time it takes for the projectile to return to its starting height.
Pleas hur

Explanation:

vi = 19m/s

vf = 0

a = - 9.81

t = ?

Notice the minus sign on the acceleration. Either Vi or the acceleration has to be negative because they go in opposite directions.

equation

a = (vf - vi)/t

Solution

- 9.81 = (0 - 19)/t     Multiply both sides by t

-9.81 * t = - 19.0     Divide both sides by 9.81

t = -19.0/9.81

t = 1.94 seconds

But that is the peak of the flight. It takes the same amount of time to descend and hit the ground.

Total time = 2 * 1.94

Total time = 3.88  seconds.

There's another way that you could do this.

Givens

vi = 19 m/s

vf = - 19 m/s

a = - 9.81

t = ?

a = (vf - vi)/t

-9.81 = (-19 - 19) / t

-9.81 * t = - 38

t = - 38/-9.81

t = - 3.87

What you are saying when you do the problem this way, is that what goes up must come down. It assumes that the landing speed is the same direction as the acceleration. It has the virtue of being shorter.

answer: the correct answer id option(d).

explanation:

snell's law refraction:

the angle of refraction =

the angle of incidence =

refractive index of air = = 1

refractive index of glycerol = 1.47

from the given options, the most closest answer to 63.72° is 65° of option (d).