Question

A player throws a ball upwards with an initial speed of 294 ms _1 . To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g=98 m s _2 and neglect air resistance)
pls answer fast​

Answer 1

Answer:

Initial velocity of the ball, u = 29.4 m/s

Final velocity of the ball, v = 0 (At maximum height, the velocity of the ball becomes zero)

Acceleration, a =g=9.8m/s  

2

 

From third equation of motion, height (s) can be calculated as:

v  

2

−u  

2

=2gs

s=(v  

2  −u  

2  )/2g= ((0)  

2  −(29.4)  

2  )/2(−9.8)=3s

Time of ascent = Time of descent  

Hence, the total time taken by the ball to return to the players hands = 3+3=6 s.