You will need some trigonometry to answer this one.
Form a triangle and recall SOH, CAH, TOA.
Since in this case the vertical component is opposite the angle, we will say that sin(20.4) = O/H (make sure your calculator is in degree mode!)
Since as I mentioned the vertical component of velocity, v_y, is opposite the angle
sin(20.4) = v_y/H
In this case the hypotenuse represents the initial velocity of the water
sin(20.4) = v_y/ 2.4 which after rearranging gives
2.4 sin(20.4) = v_y.
Rounding to 4 digits
v_y = 0.8366 m/s
For the next part we need to use kinematics
I will assume that the water will hit the beetle when the water reaches it's maximum height
Note that at the maximum height, the vertical component of the velocity will be 0.
The kinematic equation we will need is
vf^2 = vinit^2+2ad
where a is the acceleration and d is the distance. We will want to solve for d in this case. Recall we are only interested in the y -component of velocity, since at the maximum height (d) this will be vf = 0. Additionally we know the acceleration due to gravity is a = -9.8 m/s^2
setting vf = 0
0 = (2.4sin(20.4))^2 - (2*9.8*d)
0 = 0.6999 - 19.6d
19.6d = 0.6999
d = 0.6999/19.6 = 0.0357 m
this is the height of the beetle.
To determine how long the beetle has to react, we need to determine the time it took to reach this height
We will use the following kinematic equation
vf = vinit + at
but again we want vf = 0 so
0 = 0.8366 - 9.8t
9.8t = 0.8366
t = 0.8366/9.8
t = 0.0854 s
which is how long the beetle has to react.
Now note that the x-component of the velocity is v_x = 2.4*cos(20.4) = 2.2495 m/s
There is no acceleration in the x direction (of course we have been assuming this whole problem there is no air resistance).
so we can use this equation
d = vinit*t + (1/2)at^2
but since a = 0
dx = vxinit*t
and at the maximum height t = 0.0854 s
dx = 2.2495*0.0854 = 0.1921 m away in the horizontal direction.
