A rock is shot vertically upward from the edge of the top of a tall building.The rock reaches its maximum height above the top of the building 1.90 s after being shot. Then, after barely missing the edge of the building as it falls downward, the rock strikes the ground 9.00 s after it is launched. In SI units: (a) with what upward velocity is the rock shot, (b) what maximum height above the top of the building is reached by the rock, and (c) how tall is the building?
(a)
Upward motion
v=v₀ -gt
v=0
v₀ =gt =9.8•1.9 =18.62 m/s
(b)
h=v₀t -gt²/2 = 18.62•1.9 - 9.8•1.9²/2=
=35.38 – 17.69 = 17.69 m
(c)
If H is the height of the building
t₀= 9 – 1.9 = 7.1 s
H+h = gt₀²/2 = 9.8•7.1²/2 =247 m
H=247 –h= 247-17.69 =229.5 m
To solve this problem, we need to apply the equations of motion for an object in free fall. Let's break down each part of the problem.
(a) With what upward velocity is the rock shot?
To determine the rock's initial velocity, we can use the fact that the rock reaches its maximum height 1.90 seconds after being shot. At this point, the vertical velocity is zero since the rock momentarily stops before falling downward.
Using the equation of motion for vertical velocity:
v = u + at
where:
v = final velocity (0 m/s)
u = initial velocity (unknown)
a = acceleration (acceleration due to gravity, -9.8 m/s^2) Note: acceleration due to gravity is negative because it acts opposite to the direction of the upward velocity.
t = time taken (1.90 s)
Rearranging the equation:
u = (v - at)
Substituting the values:
u = (0) - (-9.8 * 1.90)
u = 18.62 m/s
So the rock is shot upward with an initial velocity of 18.62 m/s.
(b) What maximum height above the top of the building is reached by the rock?
To find the maximum height, we can use the formula for vertical displacement:
s = ut + (1/2)at^2
where:
s = displacement (unknown, maximum height)
u = initial velocity (18.62 m/s)
a = acceleration (-9.8 m/s^2)
t = time taken (1.90 s)
Rearranging the equation:
s = ut + (1/2)at^2
Substituting the values:
s = (18.62 * 1.90) + (1/2) * (-9.8) * (1.90)^2
s = 35.31 m
So the rock reaches a maximum height of 35.31 meters above the top of the building.
(c) How tall is the building?
To find the height of the building, we need to consider the total time taken by the rock from its launch until it strikes the ground.
Given the total time taken as 9.00 s, the time taken to reach the maximum height (1.90 s), and the time taken to fall back down (t + 1.90 s), we can use the equation for vertical displacement again:
s = ut + (1/2)at^2
where:
s = displacement (height of the building)
u = initial velocity (18.62 m/s)
a = acceleration (-9.8 m/s^2)
t = time taken (9 - 1.90 = 7.10 s)
Rearranging the equation:
s = ut + (1/2)at^2
Substituting the values:
s = (18.62 * 7.10) + (1/2) * (-9.8) * (7.10)^2
s = 361.61 m
So the building is approximately 361.61 meters tall.
To solve this problem, we can use the equations of motion for vertical motion.
(a) To find the initial upward velocity of the rock, we can use the equation v = u + gt, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time. Since the rock reaches its maximum height and then falls downward, we can focus on the time it takes to reach the maximum height, which is 1.90 seconds.
Using the given information, we have:
t = 1.90 s
g = 9.8 m/s^2 (acceleration due to gravity)
Using the equation v = u + gt, we can rearrange it to solve for the initial velocity (u):
u = v - gt
Since the rock reaches its maximum height, the final velocity (v) is 0 m/s at the top. Thus, the equation becomes:
u = 0 - (9.8 m/s^2)(1.90 s)
Calculating this, we get:
u = -18.62 m/s
Since the upward velocity should be positive, the initial upward velocity of the rock is approximately 18.62 m/s.
(b) To find the maximum height reached by the rock above the top of the building, we can use one of the kinematic equations for vertical motion:
s = ut + (1/2)gt^2
At the maximum height, the final displacement (s) is unknown, the initial velocity (u) is 18.62 m/s, the time (t) is 1.90 seconds, and the acceleration (g) is -9.8 m/s^2 (negative because it is acting downward).
Substituting these values into the equation, we have:
s = (18.62 m/s)(1.90 s) + (1/2)(-9.8 m/s^2)(1.90 s)^2
Calculating this value, we get:
s ≈ 17.71 m
Therefore, the maximum height above the top of the building reached by the rock is approximately 17.71 meters.
(c) To find the height of the building, we need to consider the total time it took for the rock to hit the ground after being launched. The time for the rock to reach the maximum height is 1.90 seconds, and then it falls for an additional 9.00 seconds.
The height of the building can be calculated using the equation:
s = ut + (1/2)gt^2
At the ground level, the final displacement (s) is 0 m, the initial velocity (u) is unknown, the time (t) is 9.00 seconds, and the acceleration (g) is -9.8 m/s^2 (negative because it is acting downward).
Substituting these values into the equation, we have:
0 = u(9.00 s) + (1/2)(-9.8 m/s^2)(9.00 s)^2
Simplifying this equation, we get:
-441 m = 4.9 u
Solving for u, we find:
u ≈ -90 m/s
Since the height of the building cannot be negative, we take the magnitude of the value obtained:
u = 90 m/s
Therefore, the height of the building is approximately 90 meters.