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mafiozo [28]
3 years ago
9

A body is thrown up with a velocity of 78.4 m per second.How high will it rise and how much time it will take to return to its p

oint of projection
Physics
1 answer:
a_sh-v [17]3 years ago
4 0

Answer:

The maximum height reached by the body is 313.6 m

The time to return to its point of projection is 8 s.

Explanation:

Given;

initial velocity of the body, u = 78.4 m/s

at maximum height (h) the final velocity of the body (v) = 0

The following equation is applied to determine the maximum height reached by the body;

v² = u² - 2gh

0 = u² - 2gh

2gh = u²

h = u²/2g

h = (78.4²) / (2 x 9.8)

h = 313.6 m

The time to return to its point of projection is calculated as follows;

at maximum height, the final velocity becomes the initial velocity = 0

h = v + ¹/₂gt²

h = 0 + ¹/₂gt²

h =  ¹/₂gt²

2h = gt²

t² = 2h/g

t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 313.6}{9.8} }\\\\t = 8 \ s

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The uniform rods AB and BC weigh 24 ky and kg, respectively,and the small wheel at C is of negligible weight. If the wheel ismov
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The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.

<h3>What is Potential and Kinetic energy?</h3>

Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.

mass of rod, mab = 2.4kg

mass of rod, mbc = 4kg

conservation of energy

T_{1}  + V_{1} = T_{2}  + V_{2}

h_{ab}  = h_{bc}  = 0.18m

potential energy at position 1,

V1 = m_{ab} gh_{ab}  + m_{bc} gh_{bc}

V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18

V1 = 11.30112

kinetic energy T1 at position 1 is zero

potential energy at position 2 is zero

K.E at position 2,

T_{2} = \frac{1}{2} l_{ab} w^{2}_{ab} +  \frac{1}{2} m_{bc} v^{2}_{G} +  \frac{1}{2} lw^{2}_{bc}

l_{ab} =\frac{m_{ab} l^{2}_{ab}  }{3}

= 1/3 *4 * (0.36)²

=0.10368kg m²

l =\frac{m_{bc} l^{2}_{bc}  }{12}

= 1/12 *4 * (0.6)²

=0.12kg m²

on putting the values in above equation we get,

T₂ = 1.0667vb²

0 + 11.30112 = 1.0667vb² + 0

vb = 3.2549 m/s

to learn more about Kinetic and potential energy go to - brainly.com/question/18963960

#SPJ4

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