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

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~