Answer:
23.0m/s
Explanation:
Let the initial velocity of Kyle be u
v² = u² -2gh
Where v = final velocity of Kyle.
h = height above the ground.
g = acceleration due to gravity.
At the highest point v = 0m/s
0 = u² - 2gh
u² = 2gh
Given
h = 0.538m
Taking g = 9.8m/s²
u² = 2×9.8×0.538 = 10.5448
u = 3.25m/s.
Time required to jump that high can be found from the equation
v = u - gt
0 = u - gt
u = gt
t = u/g = 3.25/9.8 = 0.332s
This time is also equal to the time required to get back down from that height to the ground.
Now considering the horizontal motion of kyle,
x = Vx t
Where Vx is the horizontal velocity
Kyle moves a horizontal distance of 0.0409m in t = 0.332s
So Vx = x/t = 0.0409/0.332 = 0.123m/s
His initial horizontal velocity is 0m/s and his final horizontal velocity is 0.123m/s. He share this velocity with the ball he catches in the air as both of them become joined together and move as one body.
The initial momentum of the ball is now shared by Kyle and the ball.
From the principle of conservation of momentum,
P1 + P2 = P
P1 = initial horizontal momentum of kyle
P2 = initial horizontal momentum of the ball
P = final momentum of the ball and Kyle
P1 = 0 (no initial horizontal velocity for kyle)
P2 = mass of ball × initial velocity of ball ub
= 0.430 × ub = 0.43ub
P = (mass of ball + mass of Kyle) × common velocity.
P = (80+0.430) × 0.123 = 9.89kgm/s
Therefore
0 + 0.43ub = 9.89
Ub = 9.89/0.123
Ub = 23m/s.
Considering the horizontal motion of Kyle movi
