A rescue team has to use a specially designed sling-shot to supply emergency medicine to a stranded hiker. The hiker's altitude above the ground is 320m and his horizontal distance from the launch site is 410m. The launching angle of the initial velocity is 68 degrees from the horizontal.

a.) what is the speed of the launch angle of 68 degrees?
b.)what should be the launch speed at the same launch angle of 68 degrees, if the person was on a small island, instead at the same level as the rescue team at the distance of 410m?

To solve this problem, we can use the equations of motion to find the speed of the launch angle of 68 degrees (a), and then substitute the values into the equations to find the launch speed at the same launch angle if the person was on a small island at the same level as the rescue team but at the distance of 410m (b).

a.) To find the speed of the launch angle of 68 degrees, we can first decompose the initial velocity into its horizontal and vertical components. The horizontal component of the velocity remains constant throughout the motion, while the vertical component is affected by gravity.

We can use the following equations:

Vertical motion:
y = y0 + (v0y)t - (1/2)gt^2
v = v0y - gt

Horizontal motion:
x = x0 + v0xt
v = v0x

Since we are given the displacement in both the horizontal and vertical directions, we can use the equation for horizontal motion to find the initial horizontal velocity, v0x:

x = v0xt
410m = v0x * t

Now, let's consider the vertical motion. The initial vertical velocity, v0y, can be found using the equation:

v = v0y - gt
0 = v0y - g * t

Here, we know that the final vertical displacement is -320m (negative due to being above the launch point) and the time taken is the same for both horizontal and vertical motion. We can substitute these values into the equation:

-320m = v0y * t - (1/2)g * t^2

Rearranging both equations, we have:

v0x = 410m / t
v0y = g * t

To find the value of time (t), we can use a trigonometric relationship. Since the launching angle is given to be 68 degrees, we can use the sine function:

sin(θ) = opposite/hypotenuse
sin(68) = v0y / v0

Remember that the hypotenuse represents the initial velocity, v0. Rearranging, we get:

v0y = v0 * sin(68)

Now we can substitute this value of v0y into our equation for v0y:

v0 * sin(68) = g * t

We can substitute the expression for v0y into our equation for the horizontal component, v0x:

410m / t = v0 * cos(68)

Now we have a system of equations:

v0 * sin(68) = g * t
410m / t = v0 * cos(68)

We can solve this system of equations simultaneously to find the speed of the launch angle.

b.) To find the launch speed at the same launch angle of 68 degrees if the person was on a small island at the same level as the rescue team but at the distance of 410m, we can use the same equations as above, but with a different horizontal displacement.

Now, the horizontal displacement is zero because the person is on the same level as the rescue team. We can solve for the initial speed, v0, using the equation:

0 = v0 * t

Since the horizontal displacement is zero, the time, t, taken to travel 410m horizontally will be the same as the time taken for the vertical displacement. Thus, we can substitute the expression for t obtained from our previous calculations into this equation to find the launch speed at the same launch angle on the small island.