Question

A placekicker kicks a football with a velocity of 35 m/sec and an angle of 25˚
above the horizontal. When the football is 10 m off the ground,

a. What is the horizontal velocity of the ball?

b. What is the vertical velocity of the ball?

c. What is the total velocity of the ball?

d. How long did it take for the ball to reach that point (10 m above the
ground)?

Answer 1

a) Vh = 35 m/s * cos(25º)
... this remains constant for the flight

c) 1/2 m v^2 = 1/2 m Vo^2 - (m g h)
... v^2 = Vo^2 - (2 g h) ... h = 10

b) Vv^2 = v^2 - Vh^2

d) h = -1/2 g t^2 + Vo sin(25º) t
... 10 = -4.9 t^2 + 35 sin(25º) t
solve for t ... the smaller value is going up, the larger is coming down

Answer 2

Vo = 35m/s[25o]. = Initial velocity.
Xo = 35*Cos25 = 31.7 m/s = Hor. component of initial velocity.
Yo = 35*sin25 = 14.8 m/s = Ver. component of initial velocity.

a. X = Xo = 31.7 m/s and does not change.

b. Y^2 = Yo^2 + 2g*h = (14.8)^2 -19.6*10 = 23.04
Y = 4.8 m/s.

c. V = Xo + Yi = 31.7 + 4.8i = 32.1m/s[8.61o]

d. Y = Yo+g*t = 4.8 @ 10 m.
14.8 - 9.8t = 4.8
t = ?.