a football is kicked with a speed of 15m/s

Question

at an angle of 35 degree from the ground.

a. how high is the ball from the ground at t=0.400s?
b. how fast is it moving(s) after being kicked?
c. at what t-values is the ball 2m from the ground?

Answer 1

Vo = 15m/s @ 35deg.
Xo = hor. = 15cos35 = 12.29m/s.
Yo = ver. = 15sin35 = 8.60m/s.

a. h = Yo*t + 0.5g*t^2,
h = 8.60*0.4 - 4.9*(0.4)^3 = 2.66m.

b. Vo = 15m/s.

c. d = Yo*t - 4.9*t^2 = 2m,
8.60t - 4.9t^2 = 2,
-4.9t^2 + 8.6t - 2 = 0.
Tr = o.27594s. = Time to rise to 2m above ground. Use Quadratic Formula and choose shortest time.

hmax = (Yf^2 - Yo^2) / 2g,
hmax=(0 - (8.60)^2) / -19.6 = 3.77m.

d = Yo*t + 4.9t^2 = 3.77 - 2 = 1.77m.
0 + 4.9t^2 = 1.77,
t^2 = 0.361,
Tf = 0.60s = Time to fall to 2m. above ground.