A football is kicked from ground level with an initial velocity of 23.4 m/s at angle of 33.5° above the horizontal. How long, in seconds, is the football in the air before it hits the ground? Ignore air resistance.

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

= 23.4m/s x sin 33.5° = 23.4 x 0.871

= 20.38m/s

by the help of kinematic equation i.e d = 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

you get 2.3=1.21*\frac{a}2 or 4.6=1.21a or a=3.802 (b)

explanation: