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Serga [27]
3 years ago
13

You throw a ball straight up. You are extra strong feeling today. It takes 11 seconds for the ball to come back down.

Physics
1 answer:
ollegr [7]3 years ago
5 0

Answer:

A. 148.23 m

B. 2.75 m/s

Explanation:

The following data were obtained from the question:

Time of flight (T) = 11 s

Maximum height (h) =?

Initial velocity (u) =?

Next, we shall determine the time taken for the ball to get to the maximum height. This can be obtained as follow:

Time of flight (T) = 11 s

Time (t) to reach the maximum height =.?

T = 2t

11 = 2t

Divide both side by 2

t = 11/2

t = 5.5 s

NOTE: Time to reach the maximum height is the same as the time taken for the ball to fall back to the plane of projection.

A. Determination of the maximum height to which the ball was thrown.

Time (t) to reach maximum height = 5.5 s

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (h) =?

h = ½gt²

h = ½ × 9.8 × 5.5²

h = 4.9 × 30.25

h = 148.23 m

B. Determination of the initial velocity.

Maximum height (h) reached = 148.23 m

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =?

u² = h/2g

u² = 148.23 / (2 × 9.8)

u² = 148.23 / 19.6

Take the square root of both side

u = √(148.23 / 19.6)

u = 2.75 m/s

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A merry-go-round rotates from rest with an angular acceleration of 1.45 rad/s2. How long does it take to rotate through (a) the
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Answer:

(a) t = 5.66 s

(b) t = 8 s

Explanation:

(a)

Here we will use 2nd equation of motion for angular motion:

θ = ωi t + (1/2)∝t²

where,

θ = Angular Displacement = (3.7 rev)(2π rad/1 rev) = 23.25 rad

ωi = initial angular speed = 0 rad/s

t = time = ?

∝ = angular acceleration = 1.45 rad/s²

Therefore,

23.25 rad = (0 rad/s)(t) + (1/2)(1.45 rad/s²)t²

t² = (23.25 rad)(2)/(1.45 rad/s²)

t = √(32.06 s²)

<u>t = 5.66 s</u>

<u></u>

(b)For next 3.7 rev

θ = ωi t + (1/2)∝t²

where,

θ = Angular Displacement = (3.7 rev + 3.7 rev)(2π rad/1 rev) = 46.5 rad

ωi = initial angular speed = 0 rad/s

t = time = ?

∝ = angular acceleration = 1.45 rad/s²

Therefore,

46.5 rad = (0 rad/s)(t) + (1/2)(1.45 rad/s²)t²

t² = (46.5 rad)(2)/(1.45 rad/s²)

t = √(64.13 s²)

<u>t = 8 s</u>

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An object in circular motion has velocity that is constantly changing. The direction of the acceleration is
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<h2>Answer: Toward the center of the circle.</h2>

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However, in this movement the velocity has a constant magnitude, but its direction varies continuously.

Let's say \vec{V} is the velocity vector, whose direction is perpendicular to the radius r of the trajectory, therefore   the acceleration \vec{a} is directed toward the center of the circumference.

 

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