a)
Y₀ = initial position of the stone at the time of launch = 0 m
Y = final position of stone = 20.0 meters
a = acceleration = - 9.8 m/s²
v₀ = initial speed of stone at the time of launch = 30.0 m/s
v = final speed = ?
Using the equation
v² = v₀² + 2 a (Y - Y₀)
inserting the values
v² = 30² + 2 (- 9.8) (20 - 0)
v = 22.5 m/s
b)
Y₀ = initial position of the stone at the time of launch = 0 m
Y = maximum height gained
a = acceleration = - 9.8 m/s²
v₀ = initial speed of stone at the time of launch = 30.0 m/s
v = final speed = 0 m/s
Using the equation
v² = v₀² + 2 a (Y - Y₀)
inserting the values
0² = 30² + 2 (- 9.8) (Y - 0)
Y = 46 m
Answer:
4.02 km/hr
Explanation:
5 km/hr = 1.39 m/s
The swimmer's speed relative to the ground must have the same direction as line AC.
The vertical component of the velocity is:
uᵧ = us cos 45
uᵧ = √2/2 us
The horizontal component of the velocity is:
uₓ = 1.39 − us sin 45
uₓ = 1.39 − √2/2 us
Writing a proportion:
uₓ / uᵧ = 121 / 159
(1.39 − √2/2 us) / (√2/2 us) = 121 / 159
Cross multiply and solve:
159 (1.39 − √2/2 us) = 121 (√2/2 us)
220.8 − 79.5√2 us = 60.5√2 us
220.8 = 140√2 us
us = 1.115
The swimmer's speed is 1.115 m/s, or 4.02 km/hr.
The electrostatic force between two charges is given by Coulomb's law:

where
ke is the Coulomb's constant
q1 is the first charge
q2 is the second charge
r is the separation between the two charges
By substituting the data of the problem into the equation, we can find the magnitude of the force between the two charges:
What is an example of how you can use scientific inquiry to solve a real life problem.