A daredevil is shot out of a cannon at 44.9◦ to the horizontal with an initial speed of 27.5 m/s. A net is positioned at a horizontal dis- tance of 41.2 m from the cannon from which the daredevil is shot.

The acceleration of gravity is 9.81 m/s2 .
At what height above the cannon’s mouth should the net be placed in order to catch the daredevil?
Answer in units of m.

H(x)=vi*t - 1/2 9.8 t^2

x=41.2 vi=27.5*sin44.9

now for t:
d(x)=27.5*cos44.9*t
but d(x)=41.2 solve for t, put it int the first equation, then solve for h(x).

To solve this problem, we can break it down into several steps:

Step 1: Determine the time it takes for the daredevil to reach the net.
To do this, we need to calculate the horizontal component of the daredevil's initial velocity. This can be found using the equation:
Vx = Vo * cos(θ)

where Vx is the horizontal component of the velocity, Vo is the initial velocity, and θ is the launch angle.

Plugging in the given values:
Vx = 27.5 m/s * cos(44.9°)
Vx = 19.638 m/s

To find the time it takes for the daredevil to reach the net, we divide the horizontal distance by the horizontal velocity:
t = d / Vx
t = 41.2 m / 19.638 m/s
t ≈ 2.099 s

Step 2: Calculate the vertical distance traveled by the daredevil during this time.
Using the vertical component of the initial velocity, we can find the distance travelled using the equation:
d = Vit + 0.5gt^2

where d is the vertical displacement, Vi is the initial vertical velocity (Viy), g is the acceleration due to gravity (9.81 m/s^2), and t is the time.

Since the daredevil is shot at an angle, we need to find the vertical component of the initial velocity (Viy) using the equation:
Viy = Vo * sin(θ)

Plugging in the values:
Viy = 27.5 m/s * sin(44.9°)
Viy ≈ 19.003 m/s

Now we can calculate the vertical displacement:
d = (19.003 m/s)(2.099 s) + 0.5(9.81 m/s^2)(2.099 s)^2
d ≈ 20.058 m

So, the daredevil will reach a height of approximately 20.058 meters above the cannon's mouth.