Answer:
Explanation:
initial total speed of the ball is given as
\(v_b = 32 m/s\)
minimum speed during its trajectory
\(v_a = 16 m/s\)
now the speed in x direction will be the minimum speed
\(v_x = 16 m/s\)
also we know that
\(v_y^2 + v_x^2 = 32^2\)
\(v_y^2 + 16^2 = 32^2\)
\(v_y = 27.7 m/s\)
now total time of the flight
\(T = \frac{2v_y}{g}\)
\(T = \frac{2(27.7)}{9.8} = 5.65 s\)
now horizontal range will be
\(R = 16(5.65) = 90.5 m\)
Part b)
as we know that at maximum height vertical velocity is zero
so we will have
\(v_{fy}^2 - v_{iy}^2 = 2as\)
now we have
\(0 - 27.7^2 = 2(-9.8)(H)\)
\(H = 39.15 m\)
