A football kicker is trying to make a field goal from 42.0m. If he kicks at 35∘ above the horizontal, what minimum speed is needed to clear the 3.10m high crossbar?

Answer 1

Vy(f)^2=Vy(i)^^+2gd

Where

Vy(f)=0m/s
Vy(i)=?
g=9.8m/s^2
d=3.10

Solve for Vy(i)

0=Vy(i)^2+2(-9.8)(3.10m)

Vy(i)^2=60.8m

Vy(i)=sqrt*[60.8m]

Vy(i)=7.79m/s

This is the velocity in the y-direction:

V(i)*Sin35º=7.79m/s

Solve for V(i):

V(i)=7.79m/s/Sin35º

V(i)=13.6m/s

Answer 2

Equation of trajectory:
y = (g x^2 sec^2(θ))/(2 v^2) + x tan(θ)

So, given:
θ = 35°
x = 42.0
y = 3.10

Find v