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kirill [66]
2 years ago
11

22. A ball is thrown horizontally from the roof of a building 12 m tall with a speed of 3.1 m/s.

Physics
1 answer:
zysi [14]2 years ago
5 0

Answer:

a) t = 1.6 s

b) d = 4.9 m

c) v = 16 m/s

d) θ = 79°

Explanation:

time of fall

t = √(2h/g) = √(2(12)/9.8) = 1.5649... s

d = vt = 3.1(1.56) = 4.8512...

vertical velocity vy = at = 9.8(1.56) = 15.336... m/s

v = √(15.336² + 3.1²) = 15.6464... m/s

θ = arctan(15.336/3.1) = 78.5724...°

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ankoles [38]

Answer:

a)  9.8 m/s

b) 4.9 m

Explanation:

This problem is a good example of Vertical motion, where the main equations for this situation are:  

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V^{2}={V_{o}}^{2}-2gy (2)  

Where:  

y is the height of the ball at a given time

y_{o}=0m is the initial height of the ball (assuming the hand of the thrower the origin of the system)  

V_{o} is the initial velocity of the ball

V is the final velocity of the ball

t=2s is the time it takes for the ball to make the complete movement (from the moment it is thrown until it falls back into the pitcher's hands)

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Knowing this, let's begin with the answers:

<h3>a) Initial velocity </h3>

In order to find the initial velocity V_{o} of the ball, we will use equation (1) and t=2s, taking into account that y=0 m and y_{o}=0m at this given time:

0=0+V_{o}t-\frac{1}{2}gt^{2} (3)  

Isolating V_{o}:

V_{o}=\frac{1}{2}gt (4)  

V_{o}=\frac{1}{2}(9.8 m/s^{2})(2 s) (5)  

Then:

V_{o}=9.8 m/s (6)  

<h3>b) Maximum height </h3>

In this part, we will use equation (2), knowing the value of the height is maximum when V=0. So, we will name this height as y_{max}:

0={V_{o}}^{2}-2gy_{max} (7)  

Isolating y_{max}:

y_{max}=\frac{{V_{o}}^{2}}{2g} (8)  

y_{max}=\frac{{(9.8 m/s)}^{2}}{2(9.8 m/s^{2})} (9)  

Finally:

y_{max}=4.9 m

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Answer:

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eimsori [14]

Answer:

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3 years ago
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