Maria throws an apple vertically upward from a height of 1.3m unit an initial velocity of velocity of 12.4m/s
1. will the apple reach a friend in a tree house 5.3m above the ground.
2. If the apple is not caught, how long will the apple be in the air before it hits the ground.
To answer these questions, we can use the equations of motion and some basic kinematics principles. Let's break down each question.
1. Will the apple reach a friend in a tree house 5.3m above the ground?
To solve this, we need to determine the maximum height the apple will reach. We can use the equation for vertical motion:
vf^2 = vi^2 + 2aΔy
Where:
vf = final velocity (0 m/s since the apple reaches its maximum height)
vi = initial velocity (12.4 m/s)
a = acceleration (for objects thrown vertically, the acceleration is gravity, which is approximately -9.8 m/s^2)
Δy = change in vertical position (maximum height - initial height)
Rearranging the equation, we get:
Δy = (vf^2 - vi^2) / (-2a)
Plugging in the values:
Δy = (0 - (12.4 m/s)^2) / (-2 * (-9.8 m/s^2))
= (0 - 153.76) / 19.6
= -7.86 m
Since the calculated value is negative, it means that the maximum height reached by the apple is below the initial height of 1.3 m. Therefore, the apple will not reach a friend in a tree house 5.3 m above the ground.
2. If the apple is not caught, how long will the apple be in the air before it hits the ground?
To determine the time it takes for the apple to hit the ground, we can use another equation of motion:
Δy = vi * t + 0.5 * a * t^2
Where:
Δy = change in vertical position (negative since the apple is falling, Δy = -1.3 m)
vi = initial velocity (12.4 m/s)
a = acceleration (gravity, -9.8 m/s^2)
t = time
Plugging in the values and rearranging the equation, we have:
-1.3 m = (12.4 m/s) * t + 0.5 * (-9.8 m/s^2) * t^2
This equation is a quadratic equation in terms of t, which we can solve using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = 0.5 * (-9.8 m/s^2), b = 12.4 m/s, and c = -1.3 m.
Calculating the values under the square root:
b^2 - 4ac = (12.4 m/s)^2 - 4 * 0.5 * (-9.8 m/s^2) * (-1.3 m)
= 153.76 - 25.48
= 128.28
Using the quadratic formula:
t = (-12.4 m/s ± √(128.28)) / (2 * 0.5 * (-9.8 m/s^2))
= (-12.4 m/s ± √(128.28)) / (-9.8 m/s^2)
Calculating the two possible values for t, we get:
t ≈ 2.1 s (approximately) or t ≈ 0.21 s (approximately)
Since we are interested in the time when the apple hits the ground, we consider the positive value of t, which is approximately 2.1 seconds. Therefore, the apple will be in the air for about 2.1 seconds before hitting the ground.