I need help, for my child. A jumper leaves the ground at an angle 32º?

This motion has two components, vertical motion which includes acceleration of gravity, and horizontal motion which is only distance and velocity. You can arrage the equations many ways and sometimes you will have two or three equations and unknowns.

http://en.wikipedia.org/wiki/Equations_of_motion

In your case you have an angle which relates the horizontal and vertical components. The total velocity at takeoff is along the 32 deg angle. So its components are Vx (horizontal velocity) and Vy (vertical velocity). And you are given the distance. So what you dont know is the time used to travel the distance. Looking over the various forms of the motion equations, I see one that relates velocity, time and distance

s = vt -1/2at2 where s is distance and t2 is t squared

in the horizonatal x direction the acceleration a is zero so

s = vt

now we realize we jump up and then fall down so the problem should be analyzed from jump to the highpoint in the air where the horzonatal distance sx is (1/2)(7.8) = 3.9 m so:

3.9 = vx(t) or t = 3.9/vx

and in the vertical direction

sy = vy(t) + (1/2)at2

where a is gravity which we know, dont know s or v (initial velocity) or t

but we know from the angle that s (the hypotenuze or diagonal traveled halfway) is

s = 3.9/cos(32) and so sy = s(sin(32)) = sin(32)(3.9/cos(32)) = 2.44 meters

so now we can make another equation for the vertical motion and isolate the time:

2.44 = vy(t) + (1/2)at2 and substitute the horizontal eq from above:

2.44 = vy(3.9/vx) + (1/2)a(3.9/vx)^2

and you also have the angle for the velocity components

vx = vcos(32)

vy = vsin(32)

Solve for v.

Use the projectile distance formula which is

Distance = (v^2 / 2) * (Sin (2 theta)

7.8 = (v^2 / 2) * sin 64 degrees

7.8 = (v^2 / 2) * 0.899

v^2 = 2 * (7.8 / 0.899)

v^2 = 14.024

v = 3.745 m/s

Please be advised there is one lacking all 52 cards who might answer your question and indicate this is the incorrect formula. Just look it up on Google, if he answers. He's done it before.

TexMav