Answer:
1 ) 3.3 [s]; 2) 49 [m]; 3) 23[m/s]
Explanation:
1)
In this problem we can use the kinematic equations, first we must find the final velocity and the we can calculate the time.
[tex]v_{f}^{2} = v_{0}^{2} +2*g*y\\ where\\v_{f} =final velocity [m/s]\\v_{0}= initial velocity [m/s] = 2 [m/s]\\\\g=gravity[m/s^2] = 9.81[m/s^2]\\y=distance = 60[m}\\\\replacing:\\\\v_{f}^{2} = 2^{2} +2*9.81*(60)\\\\v_{f}=\sqrt{1181.2} \\v_{f}=34.36[m/s][/tex]
Now using the following equation, we can find the time:
[tex]v_{f}=v_{0} +g*t\\ replacing\\34.36=2+(9.81)*t\\t=\frac{34.36-2}{9.81}=3.3[s][/tex]
Note: The acceleration is positive because the gravity is acting in the same direction towards the movement.
2)
In this problem, we must identify each of the initial data and then using the ideal kinematic equation we will find the answer.
x0= the initial position = 7[m]
v0= the initial velocity = 5 [m/s]
t = time = 3 [s]
a = acceleration = 6[m/s^2]
x = the final position, after the 3 [s]
Now replacing:
[tex]x=x_{0}+v_{0}*t+\frac{(a*t^{2} )}{2} \\x=7+5*3+\frac{(6*3^{2} )}{2}\\x=49 [m]\\[/tex]
3 ) Is the same problem that problem 2, but we must use another equation.
vf=final velocity [m/s]
v0= initial velocity [m/s] = 5 [m/s]
t= time = 3 [s]
[tex]v_{f} = v_{0} + (a*t)\\v_{f} = 23 [m/s][/tex]
