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aniked [119]
3 years ago
12

1. A baseball is thrown vertically at 16.7 m/s. What is the maimum height of the baseball?

Physics
1 answer:
Anvisha [2.4K]3 years ago
7 0

Answer:

14.2 m

Explanation:

Using conservation of energy:

PE at top = KE at bottom

mgh = ½ mv²

h = v² / (2g)

h = (16.7 m/s)² / (2 × 9.8 m/s²)

h = 14.2 m

Using kinematics:

Given:

v₀ = 16.7 m/s

v = 0 m/s

a = -9.8 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (16.7 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 14.2 m

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The weight of a box is found to be 30 N. What is the approximate mass of the box?
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What evidence did wegener have to support his theory of plate tectonics?
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Answer:

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3 years ago
You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state
posledela

Answer:

a)  y₂ = 49.1 m ,    t = 1.02 s , b)   y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

            v_{y}² = v_{oy}² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

           y-yo = v_{oy}² /2 g
            y- 0 = 10.0²/2 9.8
            y - 0 = 5.10 m
            
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
             y₂ = 5.1 + 44
             y₂ = 49.1 m
Let's use the other equation to find the time
              [tex]v_{y} = v_{oy} - g t

              t = v_{oy} / g

              t = 10 / 9.8

              t = 1.02 s

b) the maximum height

            y- 44.0 = v_{y}² / 2 g

            y - 44.0 = 5.1

            y = 5.1 +44.0

            y = 49.1 m

The time is the same because it does not depend on the initial height

              t = 1.02 s

7 0
3 years ago
A baseball is thrown with an initial velocity of 45.4 m/s at an angle of 31.2 ∘ .
sveticcg [70]

Answer:

V (initial vertical velocity) = 45.4 sin 31.2 = 23.52 m/s

1/2 m V^2 = m g h      conservation of energy

h = V^2 / (2 g) = 23.52^2 / 19.6 = 28.2 m       max height

Check:

t = 28.2 / 9.8 = 2.88 sec    time to reach max height

h = 23.52 * 2.88 - 1/2 g 2.88^2 = 27.1 m

8 0
2 years ago
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