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Licemer1 [7]
3 years ago
6

A ball is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m/s at 40

° above the horizontal. How far above or below its original level will the ball strike the opposite wall?
Physics
1 answer:
Vinvika [58]3 years ago
5 0

Answer:

Ball hit the tall building 50 m away below 10.20 m its original level

Explanation:

Horizontal speed = 20 cos40 = 15.32 m/s

Horizontal displacement = 50 m

Horizontal acceleration = 0 m/s²

Substituting in s = ut + 0.5at²

    50 = 15.32 t + 0.5 x 0 x t²

     t = 3.26 s

Now we need to find how much vertical distance ball travels in 3.26 s.

Initial vertical speed  = 20 sin40 = 12.86 m/s

Time = 3.26 s

Vertical acceleration = -9.81 m/s²

Substituting in s = ut + 0.5at²

    s = 12.86 x 3.26 + 0.5 x -9.81 x 3.26²

    s = -10.20 m

So ball hit the tall building 50 m away below 10.20 m its original level

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valentina_108 [34]

Answer:

The normal force experienced by the car is approximately 8223.2 N

Explanation:

The question relates to banking of road where the centripetal force for the circular motion of the vehicle is provided by the horizontal component of the normal reaction

The mass of the vehicle that moves around the curve, m = 900 kg

The incline of the curve, θ = 20°

The speed with which the vehicle moves around the curve, v = 12.5 m/s

The radius of the curve, R = 50 meters

We have;

N \cdot sin(\theta) = \dfrac{m \cdot v^2}{R}

Where;

θ = The angle of inclination of the road = 20°

N = The normal force experienced by the car

m = The mass of the car = 900 kg

v = The velocity with which the car is moving = 12.5 m/s

R = The radius of the curve around which the vehicle moves = 50 m

\therefore N = \dfrac{m \cdot v^2}{R \cdot sin(\theta)} = \dfrac{900 \times (12.5)^2}{50 \times sin(20^{\circ})}  = 8223.1998754586828969046217875927

The normal force experienced by the car = N ≈ 8223.2 N.

6 0
3 years ago
The height of the upper falls at Yellowstone Falls is 33 m. When the water reaches the bottom of the falls, its speed is 26 m/s.
Klio2033 [76]

Answer:

Speed of water at the top of fall = 5.40 m/s

Explanation:

We have equation of motion

v^2=u^2+2as

Here final velocity, v = 26 m/s

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a=9.8m/s^2 \\

displacement, s = 33 m

Substituting

26^2=u^2+2\times 9.8 \times 33\\\\u^2=29.2\\\\u=5.40m/s \\

Speed of water at the top of fall = 5.40 m/s

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Gnom [1K]

Answer:

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