Answer
33 m/s.
Work Step by Step
Apply equation 3-6 to the situation. We are given the launch angle. The tangent of that angle is the ratio of $v_{y0}$ to $v_{x0}$. The final ball height is 7.0 meters above the initial heigh.
Using equation 3-6, putting in y = 7.0m and the final horizontal position of 98m, we solve for the initial horizontal velocity of $v_{x0}=27.07m/s$.
The initial speed is then found by using $v_{x0}=v_0 cos36^{\circ}$ to find $v_0=33 m/s$.
