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?

###### Solution.

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 * t^{2}

where (Xo,Yo) – the initial position of the object, g = 9.81 m / s^{2} 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 .

Thereby,

X = Vo * cos(α) * t

Y = Vo * sin(α) * t – 0.5 * g * t^{2}

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 * X^{2} / (Vo^{2} * cos^{2}(α))

Now, let’s multiply by 2 * Vo^{2} * cos^{2}(α) and simplify.

The final equation will be

Vo = √( g * X^{2} / (X * sin(2α) – 2 * Y * cos^{2}(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.

Hence,

Vo = 13.58 m / s .