A 2.3kg stone is thrown from ground directly upward. the maximum height reached by the stone is 45m. Neglecting air resistance,
a) with which initial speed was the stone thrown?
b)what is the flight time of the stone until it hits the ground again?
c)what is the acceleration of the stone at the top point?
d) what is the net force acting on the stone during its free fall>
Just drag out your standard formulas of motion:
a) v = √(2gs)
b) vt - 1/2 gt^2 = 0
c) -g
d) F=mg
To find the answers to these questions, we can use the equations of motion under constant acceleration. In this case, since we are neglecting air resistance, the only force acting on the stone is gravity.
a) To find the initial speed, we can use the equation of motion:
vf^2 = vi^2 + 2ad
Since the stone reaches its maximum height and then falls back to the ground, the displacement (d) from the ground to the maximum height is half of the total displacement, which is given as 45m. Therefore, the displacement (d) for the stone's ascent is 45m/2 = 22.5m.
At the maximum height, the stone momentarily comes to rest, so its final velocity (vf) is 0. Plugging these values into the equation and solving for the initial velocity (vi), we get:
0 = vi^2 + 2(-9.8m/s^2)(22.5m)
Simplifying the equation gives us:
vi^2 = 9.8m/s^2 * 22.5m
vi^2 = 220.5m^2/s^2
Taking the square root of both sides, we find:
vi ≈ 14.85m/s
Therefore, the stone was thrown with an initial speed of approximately 14.85m/s upward.
b) To find the flight time until the stone hits the ground again, we can use the equation:
d = vit + (1/2)at^2
For the stone's ascent, the displacement is 22.5m. During the descent, the stone travels the same distance downward. So the total distance traveled is 22.5m + 22.5m = 45m.
Plugging in the values, we have:
45m = 14.85m/s * t + (1/2)(-9.8m/s^2)(t^2)
Rearranging the equation and solving for t, we get a quadratic equation in the form at^2 + bt + c = 0:
-4.9t^2 + 14.85t - 45 = 0
Solving this quadratic equation will give us the flight time (t). Using the quadratic formula, we find:
t ≈ 4.56s or t ≈ 6.44s
Since the time for ascent and descent are the same, the flight time is approximately 2 × 4.56s ≈ 9.12s.
Therefore, the flight time of the stone until it hits the ground again is approximately 9.12 seconds.
c) At the top point, the stone momentarily comes to rest, so its acceleration (a) is equal to the acceleration due to gravity, which is -9.8m/s^2 (negative because it's directed downward).
Therefore, the acceleration of the stone at the top point is approximately -9.8m/s^2.
d) During the stone's free fall, the net force acting on it is equal to the force of gravity. The force of gravity can be calculated using Newton's second law: F = ma, where F is the force, m is the mass of the stone, and a is the acceleration due to gravity.
Substituting the values, we have:
F = (2.3kg)(-9.8m/s^2)
F ≈ -22.54N
Therefore, the net force acting on the stone during its free fall is approximately -22.54 Newtons, directed downward.