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

Question

A soccer ball is kicked with an initial horizontal velocity of 18 m/s and an initial vertical velocity of 20 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 3.1 seconds after it was kicked?

m/s

6)

How high above the ground is the ball 3.1 seconds after it is kicked?

m

Explanation / Answer

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A soccer ball is kicked with an initial horizontal velocity of 18 m/s and an initial vertical velocity of 20 m/s.

a. What is the initial speed of the ball?

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

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

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

e. What is the speed of the ball 2.9 seconds after it was kicked?

f. How high above the ground is the ball 2.9 seconds after it is kicked?

Answer

a) H = v^2*sin^2(58)/2g

21-1.5 = v^2*sin^2(58)/19.6

V = 23.05m/s

b) H = h + vo^2*sin^58/2g

H = 1.5 + 21*21*sin58*sin58/19.6

H = 17.68 m

---------------------

2)

a) v = sqrt(Vx^2+Vy^2)

v = sqrt(18^2+20^2) = 26.9 m/s

---------------------------

b) tan theta = Vy/Vx

theta = tan^-1(20/18) = 48 degrees

-----------------------------

c) H = Vy^2/2g = 20*20/19.6 = 20.48 m

----------------------

d) R = v^2*sin(2*48)/g

R = 73.43 m

-----------------

e) V1y = vy - gt = 20-(9.8*2.9) = -8.42 m/s

speed v2 = sqrt(V1y^2+vx^2) = 19.87 m/s

----------------------

f) y = vy*t - 0.5*g*t^2

y = 16.791 m

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