A ball is thrown vertically and reaches a height of 25m. What was it's initial velocity?
v = Vi - g t
at top v = 0
so
0 = Vi - 9.81 t
Vi = 9.81 t
h = Hi + Vi t - 4.9 t^2
but t = Vi/9.81
so
25 = Vi (Vi/9.81) - 4.9 (Vi^2/9.81^2)
25 = Vi^2 (.102 - .051)
Vi^2 = 490
Vi = 22.1
alternatively
(1/2) m Vi^2 = m g h
or
Vi = sqrt (2 g h)
= 22.14
To find the initial velocity of the ball, we can use the kinematic equation for vertical motion:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity (which is 0 as the ball reaches its highest point)
vi = initial velocity (what we're trying to find)
a = acceleration due to gravity (-9.8 m/s^2, assuming no air resistance)
d = change in height (25m, in this case)
Rearranging the equation, we have:
vi^2 = -2ad
Plugging in the given values, we get:
vi^2 = -2(-9.8 m/s^2)(25m)
Simplifying further:
vi^2 = 490 m^2/s^2
To find vi, we take the square root of both sides of the equation:
vi = √(490 m^2/s^2)
Evaluating the square root:
vi ≈ 22.14 m/s
Therefore, the initial velocity of the ball was approximately 22.14 m/s.