Answer:
4.16 seconds
Explanation:
Data:
Acceleration a = -9.8 m/s²
Initial velocity 'vi' = 23.4m/s
as the ball traveled up and eventually came back down to the same level, the vertical distance y traveled by the football is zero.
therefore, y = 0
next is to find the vertical velocity of the football.
We need to find the speed of the ball immediately after it is kicked.
So, let it be represented by [tex]v_{iy}[/tex]
[tex]v_{iy}[/tex] = 23.4m/s x sin 33.5° = 23.4 x 0.871
[tex]v_{iy}[/tex] = 20.38m/s
by the help of kinematic equation i.e d = [tex]v_{iy}[/tex] t + 0.5* (at²) , we'll figure out time't'
The distance 'd' would be zero as it is the vertical distance.
Putting the values in above equation, it becomes
0 = 20.38*t + 0.5* (-9.8* t²)
0 = t(20.38 - 4.9t)
20.38 = 4.9t
20.38/4.9 = t
t = 4.16 seconds
Before the football hits the ground,it was in the air for 4.16 seconds
