Aball thrown vertically upward is caught by the thrower after 1.76 s. find the initial velocity of the ball. the acceleration of gravity is 9.8 m/

Answer;

  = 8.624 m/s

Explanation;

Time of flight= 4s. (0.88 s to reach highest point and 0.88 s to come back from highest point.)

Neglecting air resistance, the time of ascend is equal to time of descend for projectile.

It means the ball is at the highest point at time 0.88 s.

Using the equation;

V = Vo + at

Where, V is the final velocity which is Zero, Vo is the initial velocity ,g is the gravitational acceleration, 9.8 m/s², t is the time.

V = Vo - gt ;   g is negative because it is against the gravity .

Vo = gt

     = 9.8 m/s² × 0.88 s

     = 8.624 m/s

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s