A rock is thrown straight up at a velocity of +30. m/s from a height of 2 m above the ground. How high does it go? How long will it take to fall back to the same height from which it was thrown?
h(t) = 2 + 20t - 4.9t^2
as with all parabolas, the vertex is at t = -b/2a = 20/9.8
time down = time up
To find out how high the rock goes, we can use the kinematic equation for vertical motion:
Final velocity squared = Initial velocity squared + 2 * acceleration * displacement
Since the rock is thrown straight up, the initial velocity is +30 m/s and the final velocity at the highest point will be 0 m/s (since it momentarily stops before falling back down).
Plugging in the values into the equation:
0² = (30)² + 2 * (-9.8) * displacement
Simplifying the equation:
0 = 900 - 19.6 * displacement
19.6 * displacement = 900
displacement = 900 / 19.6
displacement ≈ 45.92 meters
Therefore, the rock reaches a height of approximately 45.92 meters.
To calculate the time it takes for the rock to fall back to the same height, we can use another kinematic equation for vertical motion:
displacement = (initial velocity * time) + (0.5 * acceleration * time²)
Since we know the displacement is 2 meters and the acceleration is -9.8 m/s² (taking gravity into account), we can solve for time.
2 = (0 * time) + (0.5 * -9.8 * time²)
Simplifying the equation:
-4.9 * time² = 2
time² = -2 / -4.9
time² ≈ 0.408
Taking the square root of both sides:
time ≈ √0.408
time ≈ 0.639 seconds
Therefore, it takes approximately 0.639 seconds for the rock to fall back to the same height from which it was thrown.
To find out how high the rock goes, we can use the concept of projectile motion. The key equation to use here is the kinematic equation for vertical displacement:
Δy = V₀y * t + (1/2) * g * t²
Where:
Δy = Vertical displacement (unknown)
V₀y = Initial vertical velocity = +30 m/s (since it is thrown upwards)
t = Time taken (unknown)
g = Acceleration due to gravity = -9.8 m/s² (negative because it acts downwards)
Since we want to find the maximum height, we need to find the time at which the vertical displacement becomes zero.
Setting Δy = 0, we have:
0 = 30 * t + (1/2) * (-9.8) * t²
Rearranging the equation, we get a quadratic equation:
4.9 * t² - 30 * t = 0
Factoring out t:
t * (4.9 * t - 30) = 0
This equation yields two solutions: t = 0 and t = 30/4.9.
Since time cannot be negative or zero in this context, the valid solution is:
t = 30/4.9 ≈ 6.12 seconds
Now we can substitute this value of t back into the kinematic equation to find the vertical displacement:
Δy = 30 * 6.12 + (1/2) * (-9.8) * (6.12)²
Δy ≈ 91.8 meters
Therefore, the rock reaches a maximum height of approximately 91.8 meters.
To find how long it takes for the rock to fall back to the same height, we can use a simpler formula. The time it takes for an object to fall from a given height can be found using the following equation:
h = (1/2) * g * t²
Where:
h = Height (unknown)
g = Acceleration due to gravity = -9.8 m/s²
t = Time taken (unknown)
In this case, the height is 2 meters. Substituting these values into the equation, we get:
2 = (1/2) * (-9.8) * t²
Rearranging and solving for t, we get:
t² = 2 / ((1/2) * (-9.8))
t² ≈ 0.408
t ≈ √0.408
t ≈ 0.639 seconds
Therefore, it takes approximately 0.639 seconds for the rock to fall back to the same height from which it was thrown.