Answer:
0.99526 s = ~ 1 s
Explanation:
As we know, S = gt or t = S/g
g = 9.8 m/s = 32.1522 ft/s
t = 32 / 32.1522 = 0.99526 s
Answer:
The time taken for the package to reach the ground is 2.71 s
Explanation:
When the package is dropped, it will first move upward due to inertia. Thus, first we calculate the time taken by it to reach the highest point:
We, have:
Initial Velocity = Vi = 32 ft/s
Final Velocity = Vf = 0 ft/s (since the package stops momentarily at highest point)
g = - 32.2 ft/s²
tup = Time taken to reach highest point
Using Newton's 1st Equation of Motion:
Vf = Vi + gt
tup = (Vf - Vi)/g
tup = (0 ft/s - 32 ft/s)/-32.2 ft/s²
tup = 0.99 sec
Now, using 2nd equation of motion, we find the height traveled upward by package:
hup = Vi t + (1/2)gt²
hup = (32 ft/s)(0.99 s) + (1/2)(-32.2 ft/s²)(0.99 s)²
hup = 31.8 ft - 15.78 ft
hup = 16 ft
Now, we find the time taken by the package to travel from highest point to ground.
Applying Newton's 2nd equation of motion from top to bottom.
tdown = tie taken by package to reach ground from highest point
Vi = 0 ft/s (since at highest point package stops)
g = 9.8 ft/s²
h = total height from ground = 32 ft + tup = 32 ft + 16 ft = 48 ft
h = Vi t + (1/2)g tdown²
48 ft = (0 ft/s)(tdown) + (1/2)(32.2 ft/s²) tdown²
tdown² = (48 ft)(2)/32.2 ft/s²
tdown = √2.98 s²
tdown = 1.72 sec
Now, the total time taken by the package to reach the ground is given as:
t = tup + tdown
t = 099 s + 1.72 s
t = 2.71 s
