A. The acceleration of the ball while it is in flight?
magnitude is 0 m/s² (magnitude is zero)
B. The velocity of the ball when it reaches its maximum height is 0 m/s (magnitude is zero)
C. The initial velocity of the ball 8.036 m/s upward
D. The maximum height reached by the ball is 3.29 m
<h3>A. How to determine the acceleration in the flight</h3>
Considering that the ball came to rest after 1.64s, it means the entire acceleration of the flight is zero as the ball was not moving in any form again.
<h3>B. How to determine the velocity at maximum height</h3>
At maximum height, the velocity of the ball is zero as it no longer has magnitude to keep going upwards. Hence the ball begins to ball down.
<h3>C. How to determine the initial velocity</h3>
- Acceleration due to gravity (g) = 9.8 m/s²
- Final velocity (v) = 0 m/s
- Time of flight (T) = 1.64 s
- Time to reach maximum height (t) = T / 2 = 1.64 / 2 = 0.82 s
- Initial velocity (u) =?
v = u - gt (since the ball is going against gravity)
0 = u - (9.8 × 0.82)
0 = u - 8.036
Collect like terms
u = 0 + 8.036
u = 8.036 m/s upward
<h3>D. How to determine the maximum height reached by the ball</h3>
- Time to reach maximum height (t) = T / 2 = 1.64 / 2 = 0.82 s
- Acceleration due to gravity (g) = 9.8 m/s²
- Maximum height (h)
h = ½gt²
h = ½ × 9.8 × 0.82²
h = 3.29 m
Learn more about motion under gravity:
brainly.com/question/20385439
#SPJ1
Hello,
The answer is option B KE=1/2mv^2.
Reason:
In order to calculate the kinetic energy of a object you need to use option B which is the correct formula to find the kinetic energy.
If you need anymore help feel free to ask me!
Hope this helps!
~Nonportrit
They do not demonstrate Earth's tilt. In fact, they're not "used" to demonstrate anything. It works the other way:. When you observe the Coriolis effect and the behavior of the Foucault pendulum, and you try to explain why the behave the way they do, one possible simple explanation for both of them is the Earth's ROTATION. Then, when you also observe the rising and setting of the sun and moon, and you also notice how the NUMBERS all go together, the case for the rotating, spherical Earth gets stronger and stronger.
Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
His model was also called the Planetary model
