Question

a ball is thrown by a player with a speed of 29.4 m/s in vertically upward direction. calculate the - a]maximum height reached
b]what is the instantaneous velocity and acceleration at the highest point
c]after how long the ball reaches the hands of a player​

Answer 1

Answer: [B]At maximum height, velocity of the ball becomes zero. Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value i.e., 9.8m/s  

2

[C] & [A]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)=3m

Time of ascent = Time of descent  

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

[A]

Answer 2

Answer:

a) first use the formula v^2-u^2/2a

so, 0^2-29.4^2/2×9.8

so, -864.36/19.6

so, maximum height it will reach is 44.1M

b)