A stone is thrown vertically upward with a speed of (a) How fast is it moving when it reaches a height of 11.0 m? (b) How long is required to reach this height? (c) Why are there two answers to (b)?
To solve this problem, we can use the equations of motion for vertical motion.
(a) To find the velocity when the stone reaches a height of 11.0 m, we can use the equation:
vf^2 = vi^2 + 2 * a * d
where:
vf = final velocity (unknown)
vi = initial velocity (given as 0 m/s because the stone is thrown vertically upward)
a = acceleration (acceleration due to gravity, -9.8 m/s^2)
d = displacement (11.0 m)
Plugging in the values into the equation, we have:
vf^2 = 0^2 + 2 * (-9.8) * 11.0
Simplifying this equation, we get:
vf^2 = -215.6
Since velocity cannot be negative in this context, we can disregard the negative sign. Thus:
vf = √215.6
vf ≈ 14.7 m/s
Therefore, the stone is moving at approximately 14.7 m/s when it reaches a height of 11.0 m.
(b) To find the time required to reach a height of 11.0 m, we can use the equation:
d = vi * t + 0.5 * a * t^2
where:
d = displacement (11.0 m)
vi = initial velocity (0 m/s)
a = acceleration (-9.8 m/s^2)
t = time (unknown)
Plugging in the values, we have:
11.0 = 0 * t + 0.5 * (-9.8) * t^2
Simplifying this equation, we get:
11.0 = -4.9 * t^2
Dividing both sides by -4.9:
-2.2449 ≈ t^2
Taking the square root:
t ≈ √(-2.2449)
t is not a real number
Since time cannot be imaginary in this context, it means that the stone will never reach a height of 11.0 m if it is thrown vertically upward.
(c) There are two answers to (b) because the equation for time is a quadratic equation, and quadratic equations can have two real roots. In this case, the equation yields non-real roots, indicating that the stone will not reach a height of 11.0 m on its way up. If we were considering the time it takes for the stone to reach this height on its way down, the quadratic equation would have two real roots, representing the time taken for the stone to go up and then come back down to the same height.
To solve these questions, we can apply the laws of motion and use the equations of motion.
(a) To find the speed of the stone when it reaches a height of 11.0 m, we can use the equation for velocity:
v = u + at
In this case, the stone is thrown vertically upward, so its initial velocity (u) is positive. The acceleration (a) is due to gravity and is negative since it acts downward. We can assume the acceleration due to gravity to be 9.8 m/s^2.
Given:
Initial velocity (u) = ? (not given)
Acceleration (a) = -9.8 m/s^2
Final velocity (v) = ?
Height (s) = 11.0 m
Since the stone is thrown vertically upward, it will eventually come to a stop at its highest point (when it reaches a height of 11.0 m). At the highest point, its final velocity (v) will be zero.
0 = u - 9.8t
Simplifying the equation, we get:
u = 9.8t
Substituting this value of u in the equation for velocity, we have:
0 = 9.8t - 9.8t
0 = 0
Therefore, the speed of the stone when it reaches a height of 11.0 m is zero.
(b) To find the time required to reach a height of 11.0 m, we can use the equation for displacement:
s = ut + (1/2)at^2
Given:
Initial velocity (u) = ? (not given)
Acceleration (a) = -9.8 m/s^2
Height (s) = 11.0 m
Time (t) = ?
At the highest point, the final displacement (s) will be zero (since the stone is at its maximum height).
0 = ut + (1/2)(-9.8)t^2
Simplifying the equation, we get:
4.9t^2 - ut = 0
This equation is a quadratic equation, which means it can have two solutions for time (t). Therefore, there are two answers to this question.
(c) The reason there are two answers to part (b) is that a projectile thrown vertically upward reaches the same height twice during its trajectory. Once during the ascent (moving upward) phase and once during the descent (moving downward) phase.
The time taken for the stone to reach a height of 11.0 m during the ascent is the same as the time taken during the descent. Therefore, there are two solutions for time (t) in this case.