What was the child's initial speed?

Question

A child runs down a 11 degree hill and then suddenly jumps upward at a 15 degree angle above horizontal and lands 1.3 meters down the hill.

Answer 1

A lot of problem can be solved algebraically.
My understanding is that this 1.3 meters were measured along the slope. If not, you may modify my equation accordingly.
Assume the child's initial jumping speed is v and in air time is t.
Thus the child's initial horizontal and vertical jumping speeds are: v*cos15° and v*sin15°, respectively. Since his horizontal speed would not be influenced by gravity or anything before hitting ground, it should be a constant. The in-air time can be calculated:
t = (1.3cos11°)/(v*cos15°)
With the in-air time, this vertical displacement is:
- 1.3sin11° = t*v*sin15° - 0.5g*t^2
= 1.3cos11°tan15° - 0.5g*(1.3cos11°)^2/(v*cos15°)^2
or: (v*cos15°)^2 = 0.5g*(1.3cos11°)^2 /(1.3cos11°tan15° + 1.3sin11°)
v can easily be solved.