Answer: the maximum heigth of the stadium at ist back wall is 151.32 ft
Explanation:
1. use the position (x) equation in parobolic movement to find the time (t)
565 ft = [frac{176 ft}{1 s\\}[/tex] * cos (35°) * t
t= 3.92 s
2. use the position (y) equation in parabolic movement to find de maximun heigth the ball reaches at 565 ft from the home plate.
y= [[frac{176 ft}{1 s\\}[/tex] * sen (35°) * 3.92 s] - 
y= 148.32 ft
3. finally add the 3 ft that exist between the home plate and the ball
148.32 ft + 3 ft = 151.32
Answer:
the free-fall acceleration on the moon is 1.68 m/s^2
Explanation:
recall the formula for the gravitational potential energy (under acceleration of gravity "g"):
PE = m * g * h
replacing with our values for the problem:
46 J = 91 * g * 0.3
solve for the "g" on the Moon:
g = 46 / (91 * 0.3)
g = 1.68 m/s^2
The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.
To find the answer, we need to know about the tension.
<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>
- Let's draw the free body diagram of the system using the given data.
- From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
- For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

- We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

- To find Ny, we need to find the tension T.
- For this, we can equate the net horizontal force.

- Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,

- Thus, the magnitude of the force that the beam exerts on the hi.nge will be,

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.
Learn more about the tension here:
brainly.com/question/28106871
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Answer:
6.32m/s
Explanation:
note:Now these calculations are based in the fact that acc. due to gravity is 10m/s²
okay so I'm thinking you think the speed of a body depends on the mass of the body also,umh... well it doesn't at all!
when two bodies of different masses fall from the same height,they fall at the same time( this is just to say)
now enough of the talking let solve....
so the ball was dropped .ie from rest to the ground through a distance of 2m,
the formula for calculating the distance if a body moving in a straight line is given by:
S=ut + ½at² where u is initial velocity, a is acceleration ( of the body or due to gravity, but since its falling freely under the influence of gravity its " we use the acceleration due to gravity ,which is 10m/s²) and t is the time taken to cover the distance.
from our question the ball was dropped from rest thus its u is 0 therefore we use this equation to find the time it took to touch ground (S=½at²)
solving ....
we get t to be 0.632s
to find the speed we substitute t in the equation below:
V=u+at ,but since u=0
V=at =10•0.632=6.32m/s
therefore the speed the body uses to strike the ground is 6.32m/s