Answer:
(a) 3.06 s
(b) 45.92 m
(c) 6.12 s
(ch) 1.43 s
Explanation:
Vertical Launch Upwards
In a vertical launch upwards, an object is launched vertically up without taking into consideration any kind of friction with the air.
If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:
[tex]\displaystyle h_m=\frac{v_o^2}{2g}[/tex]
Similarly, the time it needs to reach the maximum height is:
[tex]\displaystyle t_m=\frac{v_o}{g}[/tex]
The object's speed (vf) after time t is:
[tex]v_f=v_o-g.t[/tex]
The baseball is thrown up at vo=30 m/s.
(a) It's required to calculate the time it needs to reach the maximum height:
[tex]\displaystyle t_m=\frac{30}{9.8}=3.06[/tex]
[tex]t_m=3.06~s[/tex]
(b) The maximum height is:
[tex]\displaystyle h_m=\frac{30^2}{2(9.8)}=45.92[/tex]
[tex]h_m=45.92~m[/tex]
(c) The time needed to travel up is the same time required to return to the starting point:
[tex]t_f=2*t_m=2*3.06[/tex]
[tex]t_f=6.12~s[/tex]
(ch) The speed will be vf=16 m/s at a certain time. Using the equation
[tex]v_f=v_o-g.t[/tex]
We solve for t:
[tex]\displaystyle t=\frac{v_f-v_o}{g}[/tex]
[tex]\displaystyle t=\frac{30-16}{9.8}[/tex]
t = 1.43 s
