A long jumper takes of at an angle of 20 degrees and a speed of 11.0 m/sec. The max height achieved is 0.722 meters an the max length is 7.94 meters. At what angle and speed would the jumper have to take off to get the same max length, but twice the max height?

To find the angle and speed at which the jumper would have to take off to achieve the same max length but twice the max height, we can use the principles of projectile motion. In projectile motion, the initial velocity can be broken down into horizontal and vertical components.

Let's assume the initial speed required is represented by 'v' meters per second, and the launch angle required is represented by 'θ' degrees.

First, we need to determine the initial vertical velocity (v_y) and the initial horizontal velocity (v_x) components for the original takeoff angle of 20 degrees and speed of 11.0 m/s.

v_y = v_initial * sin(θ)
v_x = v_initial * cos(θ)

Using this information, we can find the time of flight (t) for the original jump:

t = (2 * v_y) / g

where 'g' is the gravitational acceleration (approximately 9.8 m/s^2).

Now, let's calculate the original jump's maximum height (h):

h = (v_y^2) / (2 * g)

And the original jump's maximum horizontal distance (d):

d = v_x * t

Now, to find the new angle and speed required for the same max length but twice the max height, we'll use the following steps:

1. Keeping the same max length, we'll use the original jump's horizontal distance (d) as a reference: d = 7.94 meters.

2. Twice the max height value becomes: 2 * h = 2 * 0.722 meters = 1.444 meters.

3. We can set up the following equation for the new jump:

1.444 = (v_initial * sin(θ_new))^2 / (2 * g)

Solving this equation will give us the new value for the sine of the launch angle (θ_new). We can then find the new launch angle (θ_new) itself using the inverse sine function.

4. Once we have the new θ_new, we can find the new vertical velocity (v_y_new):

v_y_new = v_initial * sin(θ_new)

5. Since we want the same max length, we'll keep the original horizontal velocity (v_x) value: v_x = v_initial * cos(20°).

6. Finally, we can find the new required initial speed (v_new) for the same max length but twice the max height:

v_new = sqrt((v_x^2) + (v_y_new^2))

Now with these steps, we can calculate the new launch angle and speed required to achieve the same max length but twice the max height of the original jump.