Question

A player throws a ball upwards with an initial speed of 29.4 ms. The time
taken by the ball to return to the player's hands is. (Take g = 9.8 ms and
neglect air resistance)
(A) 10 sec
(B) 8 sec
(C) 12 sec
(D) 6 sec​

Answer 1

Answer:

(B) 8 sec

Step-by-step explanation:

I DONT KNOW

I DONT KNOW

Answer 2

Answer:

please mark me as brainliest

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.

hope this helps u!!