A soccer player kicks a ball from ground level with an initial velocity of 25 m/

Question

A soccer player kicks a ball from ground level with an initial velocity of 25 m/s at an angle of 30o above the horizontal. In all calculations, treat gravitational acceleration as g = 10 m/s^2.

1. After 2.0 s, how many meters above the projection point is the ball?

2. After how many seconds does the ball reach its highest point?

3. What is the highest point in meters that the ball will reach?

4. How far (in meters) from the starting point will the ball land?

5. What is the ball's speed in m/s when it lands?

6. What angle does the velocity vector make with the horizontal when the ball lands?

Explanation / Answer

part a )

in y direction uy = vosin30

voy = 12.5

y = uy*t - 1/2*g*t^2

t = 2

y = 12.5*2 - 1/2*10*4

y = 25 - 20

y = 5 mt above

part 2)

at highest point vy = 0

vy = uy - gt

t = uy/g = 1.25 s

part 3 )

vy^2 = uy^2 - 2gs

vy = 0

s = uy^2/2g

s = 12.5^2 /20 = 7.8125 m

part 4 )

range = uo^2 sin2theta/g

range = 25^2 * sin2(30) / 10

range = 54.13 m

part 5 )

at landing

vx = ux

vy = -uoy

v = sqrt(vx^2 + vy^2 )

v = 25 m/s

part 6 )

tan theta = vy/uy

vy = -uoy

vx = ux

theta = -30 degree

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