The kinematic movement equations are
(x,y) = (v_0 costheta_0 t, h+v_0 sintheta_0 t -1/2g t^2)(x,y)=(v0cosθ0t,h+v0sinθ0t−12gt2)
calling (x_1,0)(x1,0) the point of impact after t_1t1 seconds we have
(x_1,0) = (v_0 costheta_0 t_1, h+v_0 sintheta_0 t_1 -1/2g t_1^2)(x1,0)=(v0cosθ0t1,h+v0sinθ0t1−12gt21)
or
{(x_1 =v_0 costheta_0 t_1 ),(0 =h+v_0 sintheta_0 t_1 -1/2g t_1^2 ):}
now solving for h, v_0 we obtain
{(v_0 = x_1/(costheta_0 t_1)),
(h =(costheta_0 g t_1^2 - 2 sintheta_0 x_1)/(2 costheta_0)):}
and then the speed just before the impact is computed using mechanical energy. So
1/2mv_0^2+h mg = 1/2 m v_1^2 and then
v_1 = sqrt(v_0^2+2hg)
The item c) is left as an exercise.
