Respuesta :
Answer:
a) v₀ =23.26 m/s
b) h= 27.6 m
c) t = 1.05 s
d) t₂= 3.69 s
Explanation:
We apply the free fall formulas:
vf= v₀+at Formula (1)
vf²=v₀²+2*a*y Formula (2)
y= v₀t+ (1/2)*a*t² Formula (3)
y:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
vf= 13 m/s
y= 19m
a=g= -9.8 m/s² = acceleration due to gravity
Problem development
a)What was its initial speed?
We apply the formula (2) for calculate initial speed (v₀):
vf²=v₀²+2*a*y Formula (2)
(13)²=v₀²+2*(-9.8)*(19)
v₀²= (13)²+372.4
[tex]v_{o} = \sqrt{(13)^{2}+372.4}[/tex]
v₀ =23.26 m/s
b)What altitude does it reach?
At maximum height (h) , vf = 0
We apply the formula (2) for calculate initial speed (h)
vf²=v₀²+2*a*y Formula (2)
0 =( 23.26)²+(2)*(-9.8)*h
19.6*h = ( 23.26)²
h = ( 23.26)² / (19.6)
h= 27.6 m
c)What time elapsed since it was thrown?
We apply the formula (1) for calculate the time since the ball was thrown until it passes through the window
vf= v₀+a*t Formula (1)
13= 23.26+(-9.8)*t
(9.8)*t = 23.26 - 13
t= 10.26/9.8
t = 1.05 s
d)What time (t₂) will it take the baseball to reach ground again, counting from the moment it passed the window upward?
We calculate the time (t) it takes for the ball to reach the maximum height since it is thrown with formula (1)
vf= v₀+a*t
0 = 23.26+(-9.8)*t
(9.8)*t = 23.26
t =( 23.26) / (9.8)
t= 2.37 s
t₁= (2.37)*(2) s = 4.74 s: total ball time going up and down
t₂= 4.74 s - 1.05 s = 3.69 s
To find the velocity or acceleration of a object , we use three equations of motion.
- a)The initial speed of the ball is 23.26 m/s.
- b)The altitude reaches by the ball is 27.6 m.
- The time elapsed since it was thrown is 1.05 s.
- The time will it take the baseball to reach ground again, counting from the moment it passed the window upward is 3.69 s.
What are the equations of motion?
To find the velocity or acceleration of a object , we use three equations of motion as,
[tex]v=u+at\\s=ut+\dfrac{1}{2}at^2\\v^2=u^2+2as[/tex]
Here, [tex]u[/tex] is the initial velocity, [tex]a[/tex] is the acceleration, [tex]v[/tex] is the velocity, [tex]t[/tex] is the time, and [tex]s[/tex] is the distance traveled.
Given information-
A baseball is seen to pass upward by a window with a vertical speed of 13 m/s.
The ball was thrown by a person 19 m below on the street.
- a)The initial speed-
[tex]13^2=v_0^2\times2\times9.81\times19\\v_0=23.26\rm m/s[/tex]
Thus the initial speed is 23.26 m/s.
- b)The altitude does it reach-
At the maximum height , the velocity of the ball becomes zero.
[tex]0=23.26^2+2\times9.81\times h\\h=27.6\rm m[/tex]
- c)The time elapsed since it was thrown-
To find the t use the formula shown above,
[tex]13=23.26+(-9.8)\times t\\t=1.05\rm s[/tex]
- d)The time will it take the baseball to reach ground again, counting from the moment it passed the window upward-
The time t has to be obtained for the maximum height as,
[tex]0=23.26\times(-9.8)t\\t=2.37[/tex]
Thus the time taken to reaches at maximum speed is 2.37, which half the time taken by the ball up and down. Thus the time for both up and down can be given as,
[tex]t=2.37\times2\\t=4.74\rm s[/tex]
Thus,
- a)The initial speed of the ball is 23.26 m/s.
- b)The altitude reaches by the ball is 27.6 m.
- The time elapsed since it was thrown is 1.05 s.
- The time will it take the baseball to reach ground again, counting from the moment it passed the window upward is 3.69 s.
Learn more about the equation of the motion here;
https://brainly.com/question/8898885
