A gymnast bounces upward from a trampoline at an initial height of 1.2m and reaches a maximum height of 4.8m before falling back down. What was his initial speed?
speed is 0 at the top so to get time to top
0 = Vi - 9.81 t
Vi = 9.81 t
then do height at that top t
4.8 = 1.2 + Vi t - 4.9 t^2
3.6 = 9.81 t^2 - 4.9 t^2
4.9 t^2 = 3.6
t = .857 second
Vi = (9.81 ) .857 = 8.41 m/s
To find the initial speed of the gymnast, we can use the principle of conservation of energy. At the highest point of the bounce, the gymnast has only potential energy, which is given by the equation:
Potential energy = mgh
where m is the mass of the gymnast, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the highest point relative to the starting point.
Since the gymnast starts with an initial height of 1.2m and reaches a maximum height of 4.8m, the height h in the equation above is (4.8 - 1.2)m = 3.6m.
At the highest point, the potential energy is equal to the kinetic energy that the gymnast had when he initially left the trampoline. The kinetic energy is given by the equation:
Kinetic energy = (1/2)mv^2
where v is the initial speed of the gymnast.
Setting the potential energy equal to the kinetic energy, we have:
mgh = (1/2)mv^2
Simplifying and canceling out the mass m, we get:
gh = (1/2)v^2
Substituting the values g = 9.8 m/s^2 and h = 3.6m, we can solve for v:
(9.8)(3.6) = (1/2)v^2
v^2 = 35.28
Taking the square root of both sides, we find:
v ≈ 5.94 m/s
Therefore, the initial speed of the gymnast was approximately 5.94 m/s.