# A soccer ball is kicked with an initial horizontal velocity of 12 m/s and an initial vertical velocity of 18 m/s.?

A soccer ball is kicked with an initial horizontal velocity of 12 m/s and an initial vertical velocity of 18 m/s.

1)What is the initial speed of the ball?=21.63

2)What is the initial angle θ of the ball with respect to the ground?=56.31

3)What is the maximum height the ball goes above the ground?

4)How far from where it was kicked will the ball land?

5)What is the speed of the ball 2.9 seconds after it was kicked?

6)How high above the ground is the ball 2.9 seconds after it is kicked?

### 1 Answer

- FiremanLv 73 weeks agoFavorite Answer
Given Ux = 12 m/s & Uy = 18 m/s

1) By u = √[(Ux)^2 + (Uy)^2] = √[(12)^2 + (18)^2] = 21.63 m/s

2) tanθ = Uy/Ux = 18/12 = 1.5 = tan56.31*

=>θ = 56.31*

3)by H = (usinθ)^2/2g = (Uy)^2/2g

=>H = (18)^2/(2 x 9.8) = 16.53 m

4)by R = u^2sin2θ/g = [(21.63)^2 x sin(2 x 56.31)*]/9.8 = 43.25 m

5)by v = u - gt

=>Vy = Uy - gt

=>Vy = 18 - 9.8 x 2.9

=>Vy = -10.42 m/s {-ve just indicating the direction is opposite to the initial velocity i.e. downward}

Thus by v = √[(Vx)^2 + (Vy)^2] = √[(12)^2 + (10.42)^2] {as Vx = Ux = constant}

=>v = 15.89 m/s

6)By s = ut -1/2gt^2

=>h = Uy x t - 1/2gt^2

=>h = 18 x 2.9 - 1/2 x 9.8 x (2.9)^2

=>h = 10.99 m