A. What to be the best height reached by the ball? assume it is kicked native ground level.

You are watching: A football is kicked straight up into the air; it hits the ground 4.6 s later.

B. V what rate did it leaving the kicker's foot?

How do i deal with these? i dont desire just response i want the process so i can memorize it

A football is kicked directly up into the air; it access time the floor 5.2s later.When an object moves upward, that is velocity decreases 9.8 m/s every second, until the velocity is 0 m/s. When the velocity = 0 m/s, the object is in ~ its greatest position. Climate the thing falls back to its initial position. As it falls, its velocity boosts 9.8 m/s every second.I always use the down trip, due to the fact that the early stage velocity – 0 m/s!!!Initial velocity increase = final velocity downThe distance up = street downThe time increase = time downTotal time = time up + time down = 5.2 sTime under = ½ * full timeTime under = ½ * 5.2 s = 2.6 sTime down =2.6 sDistance = initial velocity * time + (½ * acceleration * time ^2)Initial velocity for down expedition = 0 m/sAcceleration = 9.8 m/sTime = 2.6 sDistance = 0 * time + (½ * 9.8 * 2.6 ^2) = 33.124 mFinal velocity under = early velocity + (acceleration * time)Final velocity down = 0 + (9.8 * 2.6) = 25.48 m/sInitial velocity up = 25.48 m/sA. What to be the greatest height reached by the ball? assume it is kicked native ground level. 33.124 mB. With what speed did it leave the kicker's foot? 25.48 m/s

A. In theoretical physics, that is theorized the it takes precisely the very same amount of time to go under as the went up. Therefore we can divide by 2 to number out the time it took because that the football to with maximum height.

T = 5.2 / 2 = 2.6 s

Vf = by means of + AT

Vi = Vf - AT

Vi = (0) - (-9.8)(2.6) (Vf is zero since the ball has actually zero velocity in ~ max height)

Vi = 25.48

Vf^2 - Vi^2 = 2ΔYA

ΔY = (Vf^2 - Vi^2) / 2A

ΔY = ((0)^2 - (25.48)^2) / ((2)(-9.8))

ΔY = 33.12 m

B. We already solved because that the balls initial velocity in stimulate to reach its best height. The answer was 25.4 m/s

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A

5.2/2=2.6 this is the time it take it the foot round to with maximum height.

Calculate the distance it drops from it's maximum height. First write down all the values you know

initial velocity(Vi)=0 (no vertical motion at the peak of it's maximum height, it "stops prior to it beginning to fall downward.

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a=9.8, this is due to gravity. T= 2.6 seconds. We desire to discover d, distance. Use a kinematic equation will certainly these factors in it. In this case. D=Vit+0.5at^2

because Vi=0 Vit=0 therefor in this case d=0.5at^2

Substitute in the values. D=0.5x9.8x2.6^2=33.124m

You'll more than likely want to ring this to 33m

B

You have to use the kinematic equations again for this one. Create down the components you know, and the ones you desire to uncover out.