Aprojectile is launched vertically at 100 m/s. if air resistance can be ignored, at what speed will it return to its initial level?

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 100 m/s  

Acceleration, a = -9.81 m/s²  

Final velocity, v = ?

Displacement, s = 0 m  

Substituting  

v² = u² + 2as

v² = 100² + 2 x -9.81 x 0

v² = 100²

v = ±100 m/s

+100 m/s is initial velocity and -100 m/s is final velocity.

Speed with which it return to its initial level is 100 m/s

100 m/s

Explanation:

given,

vertical velocity of the projectile = 100 m/s

When the Projectile is launched vertically it will have two-component one is the vertical component and another is a horizontal component.

The horizontal component of velocity will remain constant.

The vertical component of the velocity will be zero at the highest point is zero and the second half motion of the projectile is the mirror of the first half movement of the projectile.

When the object reaches the initial level the velocity will be the same i.e. 100 m/s.

given data:

determine the horse power (h.p) = ?

force applied to move the box(f) = 10 pounds,

box lifted from a distance         (d) = 3 feet

time taken to lift                         (t) = 10 sec,

the horse power is defined as amount of work required to move a load. the horse power is equal to 550 foot-pound per second

                  1 h.p   = 550 foot-pound/second

it can also be defined as the the load that need to be move a body for certain distance in a certain amount of time.

mathematically,

                      h.p = (force × distance)/ time

                            = (10 × 3)/ 10

                            = 30/10

                            = 3 foot -pound/sec

hence, the horse power is needed to move the box is 3 foot-pound/second

important note: 1 h.p = 550 foot-pound/second = 746 watts

no

sodium is solid at rt