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NeTakaya
1 year ago
12

xa student fires a cannonball diagonally with an initial speed of 35.0m/s. neglect drag and the initial height of the cannonball

. what is the max height
Physics
1 answer:
ICE Princess25 [194]1 year ago
8 0

2.23 seconds is the cannons' total flight time.

What is Acceleration due to gravity ?

The acceleration an object experiences as a result of gravitational force is known as acceleration due to gravity. M/s^{2} is its SI unit. Both magnitude and direction are present.

According to the information

A diagonal ball is launched at a 45° angle from the horizontal. The ball is being fired in a projectile motion. A projectile is a motion in which a body that has been discharged into space with a velocity U is allowed to fall freely while being affected by gravity.

The cannon ball's total flight time, T, can be represented as;

where T = 2Usin(theta)/g;

U is the ball's starting speed, which is 35 m/s.

The angle the ball makes with the horizontal is theta, which is equal to 45°.

The acceleration caused by gravity is g =  9.81 m/s^{2}

Incorporating the provided data into the existing formula;

T = 35sin45°/9.81

T = 35×0.7071/9.81

T = 24.74/9.81

2.52 seconds later

2.23 seconds is the cannons' total flight time.

To know more about Acceleration due to gravity

brainly.com/question/13860566

#SPJ4

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A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at
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Answer:

Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t

Explanation:

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Lets assume that at steady state the equation of y will be

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y"=-16A cos 4t-16B sin 4t

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1 (-16A cos 4t-16B sin 4t)+8( -4A sin 4t+4B cos 4t)+16(A cos 4t+ B sin 4t)=6sin4 t

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-16A+32B+16A=0  So B=0

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Homogeneous version of above equation

m^2+8m+16=0

So CF =(C_1+tC_2)e^{-2t}

So the general equation

Y=(C_1+tC_2)e^{-2t}-3/16 cos 4t

Given that t=0 Y=0 So

C_1=\dfrac{3}{16}

t=0 Y'=0 So

C_2 =\dfrac{3}{8}

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