The vertical position of the rock is given by:
[tex]y(t)=h+v_0 t -\frac{1}{2} gt^2[/tex]
where
h=75 m is the initial height
v0=20 m/s is the initial vertical velocity
g=9.81 m/s^2 is the acceleration due to gravity
The problem asks to find the time t at which the rock hits the ground, so the time t at which y(t)=0, so the equation above becomes:
[tex]75 + 20 t - 4.9t^2 =0[/tex]
Which has two solutions:
t=-2.37 s
t=6.45 s
The first solution is a negative time so it has no physical meaning, therefore the correct answer is t=6.45 s.
Answer:
At time 6.45 seconds rock will hit ground.
Explanation:
We have equation of motion , [tex]s= ut+\frac{1}{2} at^2[/tex], s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
Let us take up direction as positive, so displacement = -75 m, initial velocity = 20 m/s, acceleration = acceleration due to gravity = -9.8[tex]m/s^2[/tex]
Substituting
[tex]-75= 20*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-20t-75=0[/tex]
t = 6.45 seconds or t =-2.37 seconds
Since negative time is not possible, t = 6.45 seconds.
So at time 6.45 seconds rock will hit ground.
