Akickball is kicked straight up at a speed of 22.4 m/s how high doex it go

you may figure out the time needed for the ball to reach zero vertical velocity by dividing the initial velocity by the acceleration

t = v₀/g

where you will find that the ball takes 2.275 seconds to come to momentary rest.

you may now use

y = y₀ + v₀t + ½at²

to find that the maximum altitude above the kicker is

y = 25.4 m

or you might use another kinematic equation which does not need the time value to get the result

v² = v₀² + 2as

s = (v² - v₀²) / 2a

s = (0² - 22.3²) / 2(-9.8)

s = 25.4 m

you have two different initial velocities in your question. if the correct one is actually 22.4 m/s, then the height result will be slightly greater at 25.6 m

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

i dont get the question

step-by-step explanation: