You lean over the edge of a building that is 106 meters above the ground and throw an object straight upwards. This object takes 3 seconds to reach its maximum vertical displacement above your hand and then falls to the ground. From the moment it leaves your hand, how long does the object take to travel to the ground in seconds?

Question

You lean over the edge of a building that is 106 meters above the ground and throw an object straight upwards. This object takes 3 seconds to reach its maximum vertical displacement above your hand and then falls to the ground. From the moment it leaves your hand, how long does the object take to travel to the ground in seconds?

Answer 1

Vf = Vo + gt,

Vo = Vf - gt,
Vo = 0 - (-9.8)3 = 29.4m/s.

h = 106 + (Vf^2 - Vo^2) / 2g,
h=106 + (0 - (29.4)^2) / -19.6=150.1m
above ground.

d = Vo*t + 4.9t^2 = 150.1m,
0 + 4.9t^2 = 150.1,
t^2 = 30.63,
t(dn) = 5.53s,

T = t(up) + t(dn) = 3 + 5.53 = 8.53s in flight.

Answer 2

To calculate the total time it takes for the object to reach the ground from the moment it leaves your hand, we need to consider its journey in two separate phases: ascending and descending.

First, let's calculate the time it takes for the object to reach its maximum vertical displacement. We know that the object's initial vertical velocity is 0 since it is thrown straight upwards. We can use the formula for displacement to find the maximum height it reaches:

s = ut + (1/2)at^2

Where:
s = displacement (maximum height)
u = initial velocity
t = time
a = acceleration (in this case, acceleration due to gravity, -9.8 m/s^2)

Since we know the displacement (106 meters), the initial velocity (0 m/s), and the acceleration due to gravity (-9.8 m/s^2), we can rearrange the formula to solve for time:

t = sqrt((2s) / a)

Using the given values, we find:

t = sqrt((2 * 106) / -9.8) ≈ 4.14 seconds

Therefore, it takes approximately 4.14 seconds for the object to reach its maximum vertical displacement.

Next, we need to consider the time it takes for the object to fall from its maximum height to the ground. Since the object is falling freely, we can use the equation for displacement once again:

s = ut + (1/2)at^2

Since the object is falling downwards, the initial velocity will be negative, i.e., -0 m/s. The time taken for the descent will be the same as the time taken for the ascent, which is approximately 4.14 seconds.

Thus, the total time it takes for the object to travel from the moment it leaves your hand to the ground is approximately 4.14 + 4.14 = 8.28 seconds.