What is the minimum velocity Vo with which an object must be thrown at an angle of α = 60° to the horizontal, so that it flew over a wall height of h = 6 m, if the wall is at the distance of d = 5 m from a point of throwing?
The equations of the motion of the object are given by the formulas
X = Xo + Vo * cos(α) * t
Y = Yo + Vo * sin(α) * t – 0.5 * g * t2
where (Xo,Yo) – the initial position of the object, g = 9.81 m / s2 is an acceleration caused by gravity and t – the time it takes the object to move.
We could choose the initial position of the object as the origin such that Xo = 0 and Yo = 0 .
X = Vo * cos(α) * t
Y = Vo * sin(α) * t – 0.5 * g * t2
From the first equation
t = X / (Vo * cos(α))
Substitute t into the second equation and we will get
Y = sin(α) * X / cos(α) – 0.5 * g * X2 / (Vo2 * cos2(α))
Now, let’s multiply by 2 * Vo2 * cos2(α) and simplify.
The final equation will be
Vo = √( g * X2 / (X * sin(2α) – 2 * Y * cos2(α)) )
Substitute the values of h = 6 m, d = 5 m and α = 60° into the final equation, where h and d are Y and X respectively.
Vo = 13.58 m / s .