A rock is thrown vertically upward with a speed of 12.0 from the roof of a building that is 40.0 above the ground. Assume free fall.
In how many seconds after being thrown does the rock strike the ground?
What is the speed of the rock just before it strikes the ground?
The position of the rock is
s = (-1/2)g*t^2 + (12m/s)*t + 40
Use the quadratic formula to find t when s = 0
The speed is |-g*t + 12| those are absolute value bars and the t from the first part.
Check my set-up for this.
Your set-up for solving the problem is correct.
To find the time it takes for the rock to strike the ground, we need to solve the quadratic equation -1/2gt^2 + 12t + 40 = 0, where g is the acceleration due to gravity.
Using the quadratic formula, t = (-b ± √(b^2 - 4ac)) / 2a, we can plug in the values:
a = -1/2g
b = 12
c = 40
Substituting these values into the quadratic formula, we have:
t = (-12 ± √(12^2 - 4*(-1/2g) * 40)) / 2(-1/2g)
Simplifying further:
t = (-12 ± √(144 + 4g * 40)) / (-g)
Now, we need to determine whether we take the positive or negative square root. Since we are interested in the time it takes for the rock to reach the ground, we take the negative square root to represent the downward motion. Therefore, the equation becomes:
t = (-12 - √(144 + 4g * 40)) / (-g)
Once you calculate the value of t, you will have the time it takes for the rock to strike the ground.
To find the speed of the rock just before it strikes the ground, we can use the formula | -g * t + 12 |, where g is the acceleration due to gravity and t is the time calculated above. This equation represents the magnitude of the velocity, which is the speed. Plugging in the values, we have:
Speed = | -g * t + 12 |
Once you calculate the value of speed, you will have the speed of the rock just before it strikes the ground. Remember to take the absolute value (represented by | |) to account for the possibility of a negative speed.