A rock is thrown directly upward from the edge of the roof of a building that is 35.3 meters tall. The rock misses the building on its way down, and is observed to strike the ground 4.00 seconds after being thrown. Neglect any effects of air resistance. With what speed was the rock thrown?

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To find the initial velocity with which the rock was thrown, we can use the equations of motion. Let's break down the problem into steps:

Step 1: Determine the final velocity of the rock when it hits the ground.
Using the equation of motion for vertical displacement:
s = ut + (1/2)gt^2
where:
s = vertical displacement (distance fallen) = 35.3 meters (height of the building)
u = initial velocity (unknown)
t = time taken to fall = 4.00 seconds
g = acceleration due to gravity = 9.8 m/s^2 (neglecting air resistance)
Since the rock is hitting the ground, its final displacement is zero:
0 = ut + (1/2)gt^2

Step 2: Solve the equation for u (initial velocity).
Rearranging the equation:
ut = -(1/2)gt^2
u = -(1/2)gt
Substituting the values:
u = -(1/2) * 9.8 m/s^2 * 4.00 s

Step 3: Calculate the value of u.
u = -19.6 m/s

Since the velocity is directed upwards, the negative sign indicates that the rock was thrown upward with an initial velocity of 19.6 m/s.

To find the initial speed with which the rock was thrown, we can use the equation of motion for vertical motion:

h = ut + (1/2)gt^2

Where:
- h is the height of the rock relative to its starting point (in this case, the ground)
- u is the initial velocity (or speed) of the rock
- t is the time it takes for the rock to reach the ground
- g is the acceleration due to gravity, which is approximately 9.8 m/s^2 on Earth

We are given the following information:
- The height of the building, h = 35.3 meters
- The time it takes for the rock to strike the ground, t = 4.00 seconds
- The acceleration due to gravity, g = 9.8 m/s^2

Since the rock is thrown directly upward, its final height relative to its starting point is zero. Therefore, we can rewrite the equation as:

0 = ut + (1/2)gt^2

Simplifying this equation further, we get:

(1/2)gt^2 = ut

We need to find the initial velocity, u. Let's isolate it:

u = (1/2)gt

Now, we can substitute the known values into the equation and solve for u:

u = (1/2)(9.8 m/s^2)(4.00 s)

Calculating this, we find:

u = 19.6 m/s

Therefore, the rock was thrown with an initial speed of 19.6 meters per second.

H=ho+ut-1/2gt^2 when H=0 t=4secs 0=35.3+4u-80 4u=44.7 u=11.2m/s