a rock is thrown straight up and reaches a height at 10 meters. 1.how long was the rock in the air 2.what is the initial vvelocity of the rock?
1. V^2 == Vo + 2g*h = 0; Vo = -2g*h = 19.6*10 = 196 m/s. = Initial velocity.
V = Vo + g*Tr = 0; Tr = -Vo/g = -196/-9.8 = 20 s. = Rise time.
Tf = Tr = 20 s. Tr+Tf = 20 + 20 = 40 s. = Time in air.
2. Vo = 196 m/s(Part 1).
To find the answers, we can use the equations of motion for a freely falling object.
1. To determine how long the rock was in the air, we can use the equation for displacement:
Δy = V₀ * t + (1/2) * g * t²,
where Δy is the displacement (change in height), V₀ is the initial velocity, t is the time, and g is the acceleration due to gravity (-9.8 m/s²).
In this case, the rock is thrown straight up, so its initial height is 0 m, and it reaches a height of 10 m. We can substitute these values into the equation:
10 = 0 * t + (1/2) * (-9.8) * t².
Simplifying the equation, we get:
5t² = 10.
Dividing by 5, we find:
t² = 2.
Taking the square root of both sides, we get:
t = √2.
Therefore, the rock was in the air for approximately √2 seconds.
2. The initial velocity of the rock can be determined using the equation for velocity:
V = V₀ + g * t,
where V is the final velocity, V₀ is the initial velocity, t is the time, and g is the acceleration due to gravity (-9.8 m/s²).
In this case, the final velocity is 0 m/s when the rock reaches its maximum height. We can substitute this value into the equation:
0 = V₀ + (-9.8) * √2.
Solving for V₀, we find:
V₀ = 9.8 * √2.
Therefore, the initial velocity of the rock is approximately 13.86 m/s.
To find the answers, we need to use basic equations of motion. Let's break this down step by step:
1. To find the time the rock was in the air, we can use the equation for vertical displacement:
h = vt - (1/2)gt^2
where:
h = height (10 meters in this case)
v = initial velocity (which we need to find)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
Since the rock is thrown straight up, its final velocity at its highest point is 0 m/s. Hence, we can substitute v = 0 in the equation above, and solve for t:
10 = 0 - (1/2) * 9.8 * t^2
Rearranging the equation, we get:
9.8t^2 = 10
Divide both sides by 9.8:
t^2 = 10/9.8
t^2 ≈ 1.02
Taking the square root of both sides, we get:
t ≈ √1.02
t ≈ 1.01 seconds
Therefore, the rock was in the air for approximately 1.01 seconds.
2. To find the initial velocity of the rock, we can use the equation for velocity:
v = u + gt
where:
v = final velocity (which is 0 m/s at the highest point)
u = initial velocity (what we need to find)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time (1.01 seconds, as we found earlier)
Substituting the known values into the equation, we can solve for u:
0 = u + 9.8 * 1.01
Rearranging the equation, we get:
u = -9.8 * 1.01
u ≈ -9.90 m/s
Therefore, the initial velocity of the rock was approximately -9.90 m/s (negative indicates upward direction).
In summary:
1. The rock was in the air for approximately 1.01 seconds.
2. The initial velocity of the rock was approximately -9.90 m/s.