A rock is shot up vertically upward from the edge of the top of the building. The rock reaches its maximum height 2 s after being shot. Them, after barely missing the edge of the building as it falls downward, the rock strikes the ground 8 s after it was launched. Find
(a) upward velocity the rock was shot at;
(b) the maximum height above the building the rock reaches; and
(c) how tall is the building?
a. V = Vo + g*t = 0.
Vo - 9.8*2 = 0.
Vo = 19.6 m/s.
b. h = Vo*t + 0.5g*t^2.
g = -9.8 m/s^2.
t = 2 s.
h = ?.
c. Tr+Tf = 2 + 2 = 4 s. To fall back to
top of bldg. Tr = Rise time. Tf = Fall
time.
8-4 = 4 s. To fall from top of bldg. to
gnd.
h = Vo*t + o.5g*t^2.
Vo = 19.6 m/s.
t = 4 s.
g = 9.8 m/s^2.
h = ?
To solve this problem, we need to use the equations of motion. Let's go step by step:
(a) To find the upward velocity the rock was shot at:
We know that the time it takes for the rock to reach its maximum height is 2 seconds. At the maximum height, the velocity will be zero. Using the equation of motion, we have:
v = u + at
Where:
v = final velocity (0 m/s at the maximum height)
u = initial velocity (what we need to find)
a = acceleration due to gravity (which is approximately -9.8 m/s^2, considering that it acts downwards)
t = time taken to reach the maximum height (2 s)
Plugging in the given values, we have:
0 m/s = u + (-9.8 m/s^2)(2 s)
0 = u - 19.6
u = 19.6 m/s (upward velocity)
Therefore, the rock was shot upward with an initial velocity of 19.6 m/s.
(b) To find the maximum height above the building the rock reaches:
To find the maximum height, we need to use the equation of motion relating displacement, initial velocity, time, and acceleration:
s = ut + (1/2)at^2
At the maximum height, the final velocity is zero. Therefore, we can rewrite the equation as:
0 = ut + (1/2)at^2
Plugging in the values we know:
0 = (19.6 m/s)(2 s) + (1/2)(-9.8 m/s^2)(2 s)^2
0 = 39.2 m - 19.6 m
0 = 19.6 m
Therefore, the maximum height above the building reached by the rock is 19.6 meters.
(c) To find the height of the building:
We know that the rock strikes the ground 8 seconds after it was launched. Using the equation of motion:
s = ut + (1/2)at^2
Where:
s = height of the building (what we need to find)
u = initial velocity (19.6 m/s)
a = acceleration due to gravity (-9.8 m/s^2, considering it acts downwards)
t = time taken to strike the ground (8 s)
Plugging in the given values:
s = (19.6 m/s)(8 s) + (1/2)(-9.8 m/s^2)(8 s)^2
s = 156.8 m - 313.6 m
s = -156.8 m
Since we are dealing with heights, the negative value doesn't make sense. Therefore, we take the magnitude of the height. The height of the building is 156.8 meters.
Therefore,
(a) The upward velocity the rock was shot at is 19.6 m/s.
(b) The maximum height above the building the rock reaches is 19.6 meters.
(c) The height of the building is 156.8 meters.