a rock is directly thrown upward from the edge of a roof of a building that is 56.3m tall. the rock misses the building on its way down, and is observed to strike the ground 4 seconds after being thrown. with what speed was the rock thrown?

hf=hi+vi*t-4.9t^2

hf=0, hi=56.3, t=4.0
solve for vi

To find the initial velocity with which the rock was thrown, we can use the following kinematic equation:

v = u + at

Where:
v = Final velocity (which is 0 m/s when the rock reaches the highest point)
u = Initial velocity (unknown)
a = Acceleration due to gravity (-9.8 m/s², considering downward direction)
t = Time taken to reach the highest point (t₁) and return to the ground (t₂)

Let's break down the problem into two parts: ascent and descent.

1. Ascent:
Since the rock is thrown vertically upward and reaches its highest point, the time taken to reach the highest point can be calculated using the formula:

s = ut + (1/2)at²

Where:
s = Distance (56.3 m)
u = Initial velocity (unknown)
a = Acceleration due to gravity (-9.8 m/s², considering upward direction)
t = Time taken to reach the highest point (t₁)

Using the above equation, we can solve for t₁:

56.3 = ut₁ - (1/2)(9.8)(t₁)² ---(Equation 1)

2. Descent:
The time taken for the rock to fall back to the ground can be calculated using the formula:

s = ut + (1/2)at²

Where:
s = Distance (56.3 m)
u = Final velocity (0 m/s)
a = Acceleration due to gravity (-9.8 m/s², considering downward direction)
t = Time taken to fall to the ground (t₂)

Using the above equation, we can solve for t₂:

56.3 = 0 - (1/2)(9.8)(t₂)² ---(Equation 2)

Now, let's solve Equations 1 and 2:

From Equation 1:
56.3 = ut₁ - (4.9)(t₁)²

From Equation 2:
56.3 = -4.9(t₂)²

Rearranging Equation 2:
(t₂)² = 56.3 / 4.9

(t₂)² = 11.5

t₂ ≈ √11.5

t₂ ≈ 3.39 seconds

Now, substituting the value of t₂ into Equation 1:

56.3 = u(4) - (4.9)(4)²

56.3 + 4.9(16) = u(4)

56.3 + 78.4 = 4u

134.7 = 4u

u = 134.7 / 4

u ≈ 33.67 m/s

Therefore, the rock was thrown with an initial speed of approximately 33.67 m/s.

To find the speed at which the rock was thrown, we can use the kinematic equation which relates the final velocity (v), initial velocity (u), time (t), and acceleration (a).

The equation is: v = u + at

In this case, the rock is thrown directly upward, so the acceleration will be due to gravity (a = -9.8 m/s^2). The final velocity when the rock hits the ground will be v = 0 m/s, as the rock comes to a stop.

The time taken (t) for the rock to reach the ground is given as 4 seconds. We need to find the initial velocity (u).

Rearranging the equation, we have: u = (v - at)

Substituting the values, we get: u = (0 - (-9.8 * 4))

Simplifying, we find: u = (0 + 39.2)

Therefore, the initial velocity (speed) at which the rock was thrown is 39.2 m/s.