We can solve this problem using the equations of motion. The key equation we'll use is the equation for the vertical displacement of an object in free fall:
h = u*t - (1/2)gt^2
where: h is the vertical displacement (height) u is the initial vertical velocity (upwards) g is the acceleration due to gravity (10 m/s²) t is the time
Given that the ball is at a height of 80m at two times, with a time interval of 6s, we can set up two equations using this equation for the two instances:
For the first time: 80 = u*t1 - (1/2)gt1^2
For the second time (6s later): 80 = u*t2 - (1/2)gt2^2
Since the time interval is 6s, we have t2 = t1 + 6.
Substituting the value of t2 in terms of t1 in the second equation, we get: 80 = u*(t1 + 6) - (1/2)g(t1 + 6)^2
Now, we have two equations with two unknowns (u and t1). We can solve this system of equations to find the value of u.
Simplifying the equations, we have: 80 = ut1 - (1/2)gt1^2 80 = ut1 + 6u - (1/2)g(t1^2 + 12t1 + 36)
Combining like terms, we get: 0 = -gt1^2 - 12g*t1 - 72u
Rearranging the equation: gt1^2 + 12g*t1 + 72u = 0
Now, we can solve this quadratic equation for t1 using the quadratic formula: t1 = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = g, b = 12*g, and c = 72u.
Plugging in the values, we have: t1 = (-12g ± √((12g)^2 - 4g72u)) / (2*g)
Simplifying further: t1 = (-12 ± √(144 - 288u)) / 2
Since time cannot be negative, we take the positive root: t1 = (-12 + √(144 - 288u)) / 2
Now, we substitute the value of t1 back into the first equation to solve for u: 80 = u*t1 - (1/2)gt1^2
Plugging in the values: 80 = u * [(-12 + √(144 - 288u)) / 2] - (1/2)*g * [(-12 + √(144 - 288u)) / 2]^2
Simplifying and rearranging the equation, we get: 160 = u * (-12 + √(144 - 288u)) - g * [(-12 + √(144 - 288u))^2] / 2
Now, we can solve this equation to find the value of u. However, it's important to note that this equation is non-linear and solving it algebraically may be challenging. Numerical methods or approximation techniques may be necessary to find a precise value for u.