Once in the air, no forces act on the froghopper, so it keeps moving with the same initial horizontal speed.
This horizontal component, is the projection of the velocity vector on the horizontal direction (x-axis):
The horizontal displacement can be simply calculated as follows:
Vertical movement:
As the vertical and horizontal are independent each other (due to they are perpendicular, so there is no projection of one movement on the other), in the vertical direction, all happens as if would be a body thrown upward with a given initial vertical velocity.
This velocity can be found as the projection of the velocity vector on the vertical direction (y-axis):
Once in the air, the gravity will cause that the froghopper be slow down, till it reaches to the maximum height, where it will come momentarily to an stop.
In that moment, we can apply the following kinematic equation:
where vfy = 0, g = -9.8m/s2, hmax = 52.7 cm= 0.527 m
Replacing by the givens, we can solve for voy:
From the equation (1), we can solve for the magnitude of the initial velocity, v₀:
B)
With the value of the magnitude of the initial velocity, we can find the horizontal component vox, as follows:
In order to know the horizontal distance travelled, we need to find the time that the insect was in the air.
We can use the equation for the vertical displacement, replacing this value by 0, as follows:
Replacing by the givens, and rearranging terms, we can solve for t:
Finally, we find the horizontal displacement, as follows:
The horizontal distance covered by the froghopper was 1.33 m.
You need to find the mass of water in the pool.
Find the volume (10 x 4 x 3) = 120 m3
Water has a density of 1000g/m3,so 120 m3 = 120 x 1000 = 120 000 kg
[delta]H = 4.187 x 120 000 x 3.4 (and the units will be kJ)
You then use the heat of combustion knowing that each mole of methane
releases 891 kJ of heat so if you divide 891 into the previous answer,
you will get the number of moles of CH4