A stone (on the edge of the roof) thrown from the top of
a building is given an initial velocity of 20.0 m/s straight
upward. The building is 50.0 m high, as shown in Fig. 4.
Using tA = 0 as the time the stone leaves the thrower’s
hand at position ○A . The gravitational acceleration is 9.8
m/s
2
. 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 which the stone returns
to the height from which it was thrown;
(e) The velocity and position of the stone at t = 5.00s.
To solve this problem, we can apply the kinematic equations and principles of projectile motion. Let's solve each part of the question step-by-step:
(a) The time at which the stone reaches its maximum height.
- The stone is thrown straight upwards, so its initial velocity is positive.
- The stone will reach its maximum height when its vertical velocity becomes zero.
- Use the equation: vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
- In this case, vi = 20.0 m/s (upward), a = -9.8 m/s^2 (it is negative because it is in the opposite direction of the initial velocity).
- Substitute the values in the equation: 0 = 20.0 - 9.8t.
- Solve for t: 9.8t = 20.0, t = 20.0 / 9.8.
- The time at which the stone reaches its maximum height is approximately 2.04 seconds.
(b) The maximum height.
- To find the maximum height, we need to use the kinematic equation: vf^2 = vi^2 + 2ad.
- At the maximum height, the final velocity is 0.
- Substitute the values in the equation: 0^2 = 20.0^2 + 2(-9.8)d.
- Solve for d (the maximum height): -196d = -400, d = -400 / -196.
- The maximum height is approximately 20.41 meters.
(c) The time at which the stone returns to the height from which it was thrown.
- The time it takes for the stone to return to its initial height is twice the time it took to reach the maximum height.
- Multiply the time calculated in part (a) by 2: 2 * 2.04.
- The time at which the stone returns to the height from which it was thrown is approximately 4.08 seconds.
(d) The velocity of the stone at which it returns to the height from which it was thrown.
- Use the equation: vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
- At the time it returns to the height from which it was thrown, the time is the value calculated in part (c).
- Substitute the values in the equation: vf = 20.0 - 9.8(4.08).
- Solve for vf: vf = 20.0 - 39.9.
- The velocity of the stone at which it returns to the height from which it was thrown is approximately -19.9 m/s (downward).
(e) The velocity and position of the stone at t = 5.00s.
- For this part, we'll calculate the velocity and position separately.
- To find the velocity, we can use the equation: vf = vi + at.
- Substitute the values: vf = 20.0 - 9.8(5.00).
- Solve for vf: vf = 20.0 - 49.0.
- The velocity of the stone at t = 5.00s is -29.0 m/s (downward).
- To find the position, we can use the equation: d = vi*t + (1/2)*a*t^2.
- Substitute the values: d = 20.0(5.00) + (1/2)*(-9.8)*(5.00)^2.
- Solve for d: d = 100.0 + (-122.5).
- The position of the stone at t = 5.00s is approximately -22.5 meters.
To solve this problem, we can apply the equations of motion for an object in free fall under the influence of gravity. The key equations are:
1. Position equation:
y = y₀ + v₀yt - (1/2)gt²
2. Velocity equation:
v = v₀y - gt
3. Time equation:
v = v₀y - gt
Now, let's solve the given problems step by step:
(a) The time at which the stone reaches its maximum height:
To find the time at maximum height, we need to determine when the vertical velocity of the stone becomes zero. Using the velocity equation:
0 = 20 - 9.8t
9.8t = 20
t = 20 / 9.8
t ≈ 2.04 seconds
(b) The maximum height:
To find the maximum height, substitute the time obtained in part (a) into the position equation:
y = 50 + 20(2.04) - (1/2)(9.8)(2.04)²
y ≈ 50.20 meters
(c) The time at which the stone returns to the height from which it was thrown:
The time of descent is equal to the time of ascent. Therefore, the total time the stone is in the air is twice the time at maximum height. So,
Total time = 2 * 2.04 ≈ 4.08 seconds
(d) The velocity of the stone at which it returns to the height from which it was thrown:
To find the velocity, substitute the total time (4.08 seconds) into the velocity equation:
v = 20 - 9.8(4.08)
v ≈ -39.98 m/s (negative sign indicates that it is moving downward)
(e) The velocity and position of the stone at t = 5.00 seconds:
To find the velocity and position at t = 5.00 seconds, substitute the time into the velocity and position equations:
v = 20 - 9.8(5.00)
v ≈ -30.0 m/s
y = 50 + (20)(5.00) - (1/2)(9.8)(5.00)²
y ≈ 0 meters (at ground level)
So, at t = 5.00 seconds, the stone has a velocity of approximately -30.0 m/s and is at ground level.
It is important to note that the negative sign in the velocity indicates the direction (upward or downward) of the stone's motion.
a. at the top, vf=0
vf=vi-9.8t
b. hf=hi-1/2 g t^2
c. double time in a.
d. vf=vi-1/2 9.8 t
e.
a. 10.2 m/s t
b. 50469.8 ft
c. 104.04 v
d. 955.5 v