First, let us assign the variables
y = 0.90 m, x= 15 m, \(h_{0}\) = 2.80 m
s= required
The vertical component of the trajectory is in uniformly accelerated motion. The equation is:
\(y= v_{0,y} t+ \frac{1}{2} a t^{2} + h_{0}\), while the horizontal component is \(x= v_{0,x} t\). Also, \(v_{0,y} =0\) since the object starts from rest (with respect to the downward motion). a is negative because it is moving downwards (a = -9.81 m/s^2). Substituting,
\(0.9= 0+ \frac{1}{2} (-9.81m/ s^{2} ) ( \frac{x}{ v_{0,x} } )^{2} + h_{0}\)
\(0.9= 0+ \frac{1}{2} (-9.81m/ s^{2} ) ( \frac{15}{ v_{0,x} } )^{2} + 2.5\)
\(v_{0,x}= v_{0} =-24.1m/s\)
The magnitude of v0, which is speed (s), is equal to 24.1 m/s
