Answer:
Solving this problem using energy notes from h. physics and the following equations
Eg= mgh Ek= ½ mv^2
Eg= gravitational energy
Ek= kinetic energy
Now, since no mass is given of the ball, both equations on their own do nothing for us, except leave us scratching our heads wondering how to figure out the problem. But, since the question states, “and no air resistance,” we now know, according to the law of conservation of energy, that the energy of the two equations will equal each other because none of the energy has dissipated or left the system.
The amount of energy present during the initial phase of the woman about to throw the ball, will be present in the final phase, which will be at its highest point (according to this problem).
So now Eg=Ek.
Knowing this, we can now set the equations equal
Eg=Ek
mgh= ½ mv^2
the two m’s cancel out, making the mass of the ball insignificant and not influential; next, substitute the values we are given in the problem (22m/s),(9.8m/s^2)
m(9.8 m/s^2)h= ½ m(22 m/s)^2
(9.8 m/s^2)h= ½ (22 m/s)^2
(9.8 m/s^2)h= ½ (484 m^2/s^2)
(9.8 m/s^2)h= (242 m^2/s^2)
h= (242 m^2/s^2)/(9.8 m/s^2)
as you can see, all units that need to be cancelled out are indeed cancelled, leaving us with just m, meters, which is what height is measured in
h= 24.69 m
So the answer is 24.69 meters. Hope this helps (there are probably more efficient ways to come upon this answer, but this was the way I did it, cause this is what I know right now)
Explanation:
The kinematic equations of motion provides a description of the variables of motion
The time the ball is in the air is is approximately 5.097 s
Reason:
The direction the ball is thrown = Upwards
The speed with which the ball is thrown, u = 25 m/s
Required:
The time the ball is in the air
Solution:
The time the ball is in the air is given by the following kinematic equation of motion ;
[tex]s = u\cdot t + \dfrac{1}{2} \cdot g \cdot t^2[/tex]
Where,
s = 0
g = 9.81 m/s²
Which gives;
[tex]0 = 25\cdot t - \dfrac{1}{2} \times 9.81 \times t^2[/tex]
[tex]25\cdot t = 4.905 \times t^2[/tex]
[tex]t = \dfrac{25}{4.905} \approx 5.097[/tex]
The time the ball is in the air is t ≈ 5.097 s
Learn more about kinematic equation of motion here:
https://brainly.com/question/12867322
