For me: WASH OUR HANDS REGULARLY
For vertical motion, use the following kinematics equation:
H(t) = X + Vt + 0.5At²
H(t) is the height of the ball at any point in time t for t ≥ 0s
X is the initial height
V is the initial vertical velocity
A is the constant vertical acceleration
Given values:
X = 1.4m
V = 0m/s (starting from free fall)
A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)
Plug in these values to get H(t):
H(t) = 1.4 + 0t - 4.905t²
H(t) = 1.4 - 4.905t²
We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:
1.4 - 4.905t² = 0
4.905t² = 1.4
t² = 0.2854
t = ±0.5342s
Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))
t = 0.53s
The third equation of free fall can be applied to determine the acceleration. So that Paola's acceleration during the flight is 39.80 m/
.
Acceleration is a quantity that has a direct relationship with velocity and also inversely proportional to the time taken. It is a vector quantity.
To determine Paola's acceleration, the third equation of free fall is appropriate.
i.e 
= 
± 2as
where: V is the final velocity, U is the initial velocity, a is the acceleration, and s is the distance covered.
From the given question, s = 20.1 cm (0.201 m), U = 4.0 m/s, V = 0.
So that since Poala flies against gravity, then we have:
= - 2as
0 = 
- 2(a x 0.201)
= 16 - 0.402a
0.402a = 16
a = 
= 39.801
a = 39.80 m/
Therefore Paola's acceleration is 39.80 m/.
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Answer: Your answer is 1250J
Explanation:
K
E
=
1/2
m
v
2
The mass is
m
=
25
k
g
The velocity is v
=
10
m
s
−
1
So,
K
E
=
1
/2
x25
x
10
2^2=
1250
J
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