A soccer ball is kicked with an initial horizontal velocity of 14 m/s and an ini

Question

A soccer ball is kicked with an initial horizontal velocity of 14 m/s and an initial vertical velocity of 13 m/s.

1)

What is the initial speed of the ball?

m/s

2)

What is the initial angle of the ball with respect to the ground?

degrees

3)

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

m

4)

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

m

5)

What is the speed of the ball 1.7 seconds after it was kicked?

m/s

6)

How high above the ground is the ball 1.7 seconds after it is kicked?

m

Explanation / Answer

1) ux = 14 m/s

uy = 13 m/s

speed = sqrt(ux^2 + uy^2)

= sqrt(14^2 + 13^2) = 19.10 m/s

2) @ = tan^-1(uy / ux) = 42.88 deg

3) maximum height is when vertical velocity will become zero.

using v^2 - u^2 = 2ad

0^2 - 13^2 = 2(-9.81)H

H = 8.61 m

4) for time period using v = u + at

-13 = 13 - 9.81t

t = 2.65 s

distance = 14 x 2.65 = 37.10 m

5. after 2.7 s

vx =ux = 14 m/s

vy = 13 - 9.81(1.7) = - 3.68 m/s

speed = sqrt(14^2 + 3.68^2) = 14.47 m/s

6) h = uy*t + ay*t^2 /2

h = 13*1.7 + (-9.81 * 1.7^2 /2 )

h = 7.92 m

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