Answer:
The answer to the question is
The ball in the air for 3.51 s before it hits the ground
Explanation:
To solve the question. we list the variables
Intitial velocity = 16.0 m/s,
Initial height of ball from the ground = 2.5 m
acceleration due to gravity = 9.81 m/s²
The required relation is
v = u + at
where a = -g and g = acceleration due to gravity×
g = 9.81 m/s², v= 0 at top before coming back down
u = 16.5 m/s, thus 16.5 m/s = t × 9.8 m/s²
t = 16.5 ÷9.81 = 1.68 s
The ball height at top of its motion is =
S = ut - 0.5×g×t²
= 16.5×1.68 - 0.5×9.81×1.68²
= 13.876 m
Therefore total height 13.876 m+ 2.5 m = 16.376 m
and the time it takes bacjk down is given by
S = ut + 0.5gt² ⇒ 16.376 m = 0×t + 0.5×9.81×t²
16.376 m = t²×4.905 m/s² or t² = 3.339 s² or t = 1.827 s
Therefore total time = 1.807 s+ 1.68 s = 3.51 s
The ball in the air for 3.51 s