A stone is launched straight up by a slingshot. Its initial speed is 20.1 m/s and the stone is 1.30 m above the ground when launched. Assume g = 9.80 m/s2.
(a) How high above the ground does the stone rise?
(b) How much time elapses before the stone hits the ground?
a. hmax = ho + (Vf^2-Vo^2 / 2g.
hmax=1.3 +(0-(20.1)^2 / -19.6=21.9 m.
b. Tr = Vf-Vo) / g,
Tr = (0-20.1) / -9.8 = 2.05 s. = Rise
time or time to reach max ht.
h = Vo*t + 0.5g*t^2 = 21.9 m.
0 + 4.9t^2 = 21.9,
t^2 = 4.47,
Tf = 2.11 s.Time to fall to ground.
Tr + Tf = 2.05 + 2.11 = 4.16 s. = Time
in flight.
To solve this problem, we can use the equations of motion under constant acceleration. The key equation that we will use is:
v² = u² + 2as
Where:
v = final velocity
u = initial velocity
a = acceleration (in this case, acceleration due to gravity, which is -9.80 m/s²)
s = displacement
(a) How high above the ground does the stone rise?
Initially, the stone is 1.30 m above the ground and its upward velocity is 20.1 m/s. Since the stone eventually stops and falls back down, its final velocity is 0 m/s. We can use the equation to find the displacement:
0 = (20.1 m/s)² + 2(-9.8 m/s²)s
Rearranging the equation, we get:
(20.1 m/s)² = 2(-9.8 m/s²)s
solving for s:
s = (20.1 m/s)² / (2(-9.8 m/s²))
s = 40.404 m²/s² / -19.6 m/s²
s = -2.06 m
The negative sign indicates that the displacement is in the opposite direction of the initial displacement. Therefore, the stone rises 1.30 m - 2.06 m = -0.76 m above the ground. However, this negative displacement indicates that the stone reaches a maximum height of 0.76 m below its initial position.
(b) How much time elapses before the stone hits the ground?
To find the time it takes for the stone to hit the ground, we can use the equation:
v = u + at
Since the final velocity is 0 m/s and the initial velocity is 20.1 m/s, we can solve for t:
0 = 20.1 m/s + (-9.8 m/s²)t
Solving for t:
9.8 m/s²t = 20.1 m/s
t = 20.1 m/s / 9.8 m/s²
t ≈ 2.04 s
Therefore, it takes approximately 2.04 seconds for the stone to hit the ground.
To solve this problem, we can use the equations of motion.
(a) To find the maximum height the stone reaches, we can use the equation for displacement:
Δy = v₀y * t - (1/2) * g * t²
where:
Δy is the displacement or change in height (final height - initial height)
v₀y is the initial vertical velocity (upward velocity)
t is the time
g is the acceleration due to gravity (given as 9.80 m/s²)
Given information:
- Initial speed (v₀) = 20.1 m/s
- Initial height (y₀) = 1.30 m
- Acceleration due to gravity (g) = 9.80 m/s²
Since we want to find the maximum height, we need to find the time it takes for the stone to reach that height. At the maximum height, the vertical velocity will be zero (the stone momentarily stops before falling back down). So we can use that to find the time.
v_fy = v₀y - g * t
At max height, v_fy = 0. We can rearrange the equation to solve for t:
t = v₀y / g
Substituting the given values:
t = 20.1 m/s / 9.80 m/s²
Solving for t, we find:
t ≈ 2.05 s
Now that we have the time, we can use it to find the maximum height by substituting it back into the displacement equation:
Δy = v₀y * t - (1/2) * g * t²
Δy ≈ (20.1 m/s) * (2.05 s) - (1/2) * (9.80 m/s²) * (2.05 s)²
Δy ≈ 20.65 m
Therefore, the stone rises approximately 20.65 m above the ground.
(b) To find the time it takes for the stone to hit the ground, we can use the equation for displacement:
Δy = v₀y * t - (1/2) * g * t²
At the ground, the displacement Δy will be equal to zero. We want to find the time it takes for the stone to reach that point. So we can rearrange the equation to solve for t:
0 = v₀y * t - (1/2) * g * t²
(1/2) * g * t² = v₀y * t
(1/2) * g * t = v₀y
t = (2 * v₀y) / g
Substituting the given values:
t = (2 * 20.1 m/s) / 9.80 m/s²
Solving for t, we find:
t ≈ 4.11 s
Therefore, it takes approximately 4.11 seconds for the stone to hit the ground.