Abaseball 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)what was its initial speed?
b)what altitude does it reach?
c)what time elapsed since it was thrown?
d)what time will it take the baseball to reach ground again, counting from the moment it passed the window upward?

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

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

insects!

explanation:

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2