as you know, the height of the ball at distance x is
y(x) = tanθ x - 4.9/(v cosθ)^2 x^2
Using our data, we need to solve
42.2 tanθ - 4.9/(27.4 cosθ)^2 (42.2^2) = 3.44
42.2 tanθ - 4.9/(27.4^2 (1+tan^2θ))(42.2^2) = 3.44
I get tanθ = 1/3, but I don't get two values. At that speed the ball isn't going fast enough to cross the bar on its way up.
A football kicker can give the ball an initial speed of v_0 = 27.4 m/s. In order to score a field goal, the ball must go over the cross bar that is at a height of h = 3.44 m above the ground. If the kicker kicks the ball from a point d = 42.2 m from the goalposts, find the minimum and maximum angles at which he must kick the ball. Give the difference between the minimum and maximum.
3 answers
In fact, I get that a Vi = 67 m/s is required if the ball is to be able to go over the bar at its max height, and more than that if we expect to have two values for θ.
app really do be like that sometimes