A student throws a set of keys vertically upward to her sorority sister, who is in a window 3.80 m above. The keys are caught 1.80 s later by the sister's outstretched hand. (b) What was the velocity of the keys just before they were caught?
average velocity=3.8/1.8 m/s
but avg velociyt=(Vf+vi)/2=(vf+vf-g*1.8)
solve for vf
To find the velocity of the keys just before they were caught, we can use the equation of motion:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the keys are thrown vertically upward, so the acceleration is the acceleration due to gravity, which is approximately 9.8 m/s² (assuming no air resistance).
Given:
Initial position (s) = 0 m (since the keys are thrown from the ground)
Final position (s) = 3.80 m (since the keys are caught at a height of 3.80 m)
Time (t) = 1.80 s
Acceleration (a) = -9.8 m/s² (negative since it's acting opposite to the direction of motion)
We need to find the initial velocity (u) of the keys.
First, we can calculate the displacement (Δs) using the formula:
Δs = ut + (1/2)at²
Substituting the known values:
3.80 m = u(1.80 s) + (1/2)(-9.8 m/s²)(1.80 s)²
Solving this equation will give us the initial velocity (u).
3.80 m = 1.80u - 15.876 m
Rearranging the equation:
1.80u = 3.80 m + 15.876 m
1.80u = 19.676 m
Dividing by 1.80:
u = 10.932 m/s
Therefore, the initial velocity of the keys just before they were caught was approximately 10.932 m/s.