Projectiles - proof there is no solution

[Rxyz]

I don't think it is possible to solve.

By what height must a long jumper raise his centre of mass when making a jump of 9m, if his horizontal velocity on take-off is 8ms^-1 ? (15 marks)

I would solve this problem by

x=Vcos(theta)t y=Vsin(theta)t-1/2gt²

this can't be used because there is no angle given in the question.

or by

1/2mv²=mgh
v=root(2gh)
dx/dt=root(2gh)
dx=root(2gh) dt

don't know where to go from here

 

[Ishuda]

Ideally, the center of mass is considered a projectile (only continuing influence is the force of gravity) so we can separate the horizontal and vertical velocity. Let a subscript v indicate vertical and a subscript h indicate horizontal. We have
V_v = - g t + c
where V_v is the vertical velocity, g is the force of gravity (= 9.8 m/s/s) and c is an unknown constant. Integrating we have
D_v = - 0.5 g t² + c t
where D_v is the vertical distance and, because the vertical distance is zero at time 0, the integration constant is zero. Since the vertical distance is also zero when the jumper lands (has vertical distance zero) and we can write
D_v = (c - 0.5 g t ) t,
the time of landing is t_l = 2 c / g. The maximum height is obtained at half that time, so the maximum height D_vl is given by
D_vl = (c - 0.5 g c/g) c/g = 0.5 c² / g

For the horizontal component, we are given
V_h = 8
so, integrating, we have
D_h = 8 t
since the horizontal distance is zer0 at t=0, the integration constant is zero. So, given the time of landing above, the horizontal distance at the time of landing is
D_hl = 16 c / g = 9
Solve that for c and plug back into the D_vl equation above.

 

[Rxyz]

Thank you!