A stone thrown from the top of a building is given an initial velocity of 40.0 m/s straight upward. The building is 70.0m high, and the stone just misses the edge of the roof on its way down, determine
(A) The time at which the stone reaches its maximum height,
(B) The maximum height,
(C) The time at which the stone returns to the height from which it was thrown,
(D) The velocity of the stone at this instant
(E) The velocity and position of the stone at t = 10.00 s.
A. V = Vo + g*Tr.
0 = 40 - 9.8Tr, Tr = 4.08s = Rise time or time to reach max. ht.
B. hmax = ho + Vo*Tr + 0.5g*Tr^2.
hmax = 70 + 40*4.08 - 4.9*(4.08)^2 = 151.7 m. above gnd.
C. Tf = Tr = 4.08 s. = Fall time.
D. V = Vo + g*Tf = 0 + 9.8*4.08 = 40 m/s.
E. h = ho - 0.5g*Tf^2.
0 = 151.7 - 4.9Tf^2
4.9Tf^2 = 151.7,
Tf^2 = 31, Tf = 5.56 s. = Time to fall from max. ht. to gnd.
Tr+Tf = 4.08 + 5.56 = 9.64 s = Time in air. So the stone is on the gnd 10 s after release.
V^2 = Vo^2 + 2g*h = 0 + 19.8*151.7 = 3,003.7, V = 54.8 m/s.
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To solve this problem, we can use the equations of motion:
1. The equation for vertical displacement is given by:
h = (v₀y * t) + (0.5 * g * t²)
where h is the height, v₀y is the initial vertical velocity, g is the acceleration due to gravity (9.8 m/s²), and t is the time.
2. The equation for vertical velocity is given by:
vâ‚“ = vâ‚€x + (-g * t)
where vₓ is the final vertical velocity, v₀x is the initial vertical velocity, g is the acceleration due to gravity (9.8 m/s²), and t is the time.
(A) The time at which the stone reaches its maximum height:
At the maximum height, the vertical velocity becomes zero. Therefore, using equation (2):
0 = 40 - 9.8 * t_max
Solving for t_max gives:
t_max = 40 / 9.8 = 4.08 seconds
(B) The maximum height:
Using equation (1), we can find the maximum height:
h_max = (40 * 4.08) + (0.5 * 9.8 * 4.08²)
= 163.2 + 84.1
= 247.3 meters
Therefore, the maximum height is 247.3 meters.
(C) The time at which the stone returns to the height from which it was thrown:
Since the stone just misses the edge of the roof, it will take the same amount of time to come back down as it did to reach the maximum height. Therefore, the time at which the stone returns to the height from which it was thrown is t_return = 2 * t_max = 2 * 4.08 = 8.16 seconds.
(D) The velocity of the stone at this instant:
At the height from which it was thrown, the stone will have the same magnitude of velocity as its initial velocity but in the opposite direction. Therefore, the velocity of the stone at this instant is -40 m/s.
(E) The velocity and position of the stone at t = 10.00 s:
To determine the velocity, we can use equation (2):
v = 40 - 9.8 * t
v = 40 - 9.8 * 10
v = 40 - 98
v = -58 m/s
Therefore, the velocity of the stone at t = 10.00 s is -58 m/s.
To determine the position, we can use equation (1):
h = (40 * 10) + (0.5 * 9.8 * 10²)
h = 400 + 490
h = 890 meters
Therefore, the position of the stone at t = 10.00 s is 890 meters.
To solve this problem, we can use the equations of motion for projectile motion. Let's go step by step to find the answers.
(A) The time at which the stone reaches its maximum height:
To find the time at which the stone reaches its maximum height, we need to use the equation:
v = u + at
Where:
v = final velocity at maximum height (0 m/s because the stone momentarily stops at the maximum height)
u = initial velocity (40.0 m/s)
a = acceleration due to gravity (-9.8 m/s^2, negative because it acts downward)
Rearranging the equation, we have:
t = (v - u) / a
Substituting the values, we get:
t = (0 - 40.0) / -9.8
t = 4.08 seconds (approximately)
So, the stone reaches its maximum height at approximately 4.08 seconds.
(B) The maximum height:
To find the maximum height, we can use the equation of motion for vertical displacement:
s = ut + (1/2)at^2
At the maximum height, the velocity is 0 m/s, so the equation becomes:
s = ut + (1/2)at^2
0 = 40.0(4.08) - (1/2)(9.8)(4.08)^2
Simplifying the equation, we get:
0 = 163.2 - 83.88672
83.88672 = 163.2
s ≈ 83.9 meters
So, the maximum height reached by the stone is approximately 83.9 meters.
(C) The time at which the stone returns to the height from which it was thrown:
To find the time at which the stone returns to the height from which it was thrown, we know that the total time of flight will be twice the time it takes to reach the maximum height.
So, the time at which the stone returns to the height from which it was thrown is approximately:
2 * 4.08 = 8.16 seconds
(D) The velocity of the stone at this instant:
To find the velocity of the stone at the instant it returns to the height from which it was thrown, we can use the equation:
v = u + at
We have:
u = 40.0 m/s,
a = -9.8 m/s^2 (acceleration due to gravity),
t = 8.16 s
Substituting the values, we get:
v = 40.0 - 9.8(8.16)
v ≈ -79.9 m/s
So, the velocity of the stone at this instant is approximately -79.9 m/s (negative sign indicates downward direction).
(E) The velocity and position of the stone at t = 10.00 s:
To find the velocity and position of the stone at t = 10.00 seconds, we can use the equations of motion.
First, let's find the velocity:
v = u + at
Given:
u = 40.0 m/s,
a = -9.8 m/s^2,
t = 10.00 s
Substituting the values, we get:
v = 40.0 - 9.8(10.00)
v = -59.8 m/s
So, the velocity of the stone at t = 10.00 seconds is -59.8 m/s (negative sign indicates downward direction).
Next, let's find the position:
s = ut + (1/2)at^2
Given:
u = 40.0 m/s,
a = -9.8 m/s^2,
t = 10.00 s
Substituting the values, we get:
s = 40.0(10.00) + (1/2)(-9.8)(10.00)^2
s ≈ 200.0 - 490.0
s ≈ -290.0 meters
So, the position of the stone at t = 10.00 seconds is approximately -290.0 meters (negative sign indicates below the starting point, which is 70.0 meters above ground level).