How can we calculate the maximum height and velocity of a newspaper's trajectory thrown by a delivery boy? What about the acceleration and time it takes to reach the balcony?
a) Maximum Height Calculation
To find the maximum height of the paper's trajectory above the boy's hand, we can use the kinematic equation:
\[ v^2 = u^2 + 2gh \]
where:
\[ v = 0 \] (at maximum height)
\[ u = 11.34 \, \text{m/s} \] (initial vertical velocity)
\[ g = -9.8 \, \text{m/s}^2 \] (deceleration in vertical direction)
Substitute the values in the equation:
\[ 0^2 = (11.34)^2 + 2(-9.8)h \]
\[ h = \frac{(11.34)^2}{2 \times 9.8} = 6.5 \, \text{m} \]
Therefore, the maximum height of the paper's trajectory above the boy's hand is 6.5 m.
b) Velocity at Maximum Height
To find the velocity at maximum height, we use the same kinematic equation:
\[ v^2 = u^2 + 2gh \]
Substitute the known values:
\[ v^2 = (11.34)^2 + 2(-9.8) \times 6.5 \]
\[ v = \sqrt{43.15} = 6.57 \, \text{m/s} \]
The velocity at maximum height is 6.57 m/s.
c) Acceleration at Maximum Height
At maximum height, velocity is zero, so the acceleration is equal to acceleration due to gravity:
\[ -9.8 \, \text{m/s}^2 \]
Therefore, the acceleration at maximum height is -9.8 m/s².
d) Time of Arrival at Balcony
To determine the time taken for the paper to reach the balcony as it descends, we use the kinematic equation:
\[ s = ut + \frac{1}{2}at^2 \]
Where:
\[ s = 0.75 \, \text{m} \] (height of the balcony)
\[ u = 10.7 \, \text{m/s} \] (horizontal velocity)
\[ g = 9.8 \, \text{m/s}^2 \]
Solving for time using the quadratic formula:
\[ t = \frac{-10.7 \pm \sqrt{10.7^2 + 4 \times 0.49 \times 0.75}}{2 \times 0.49} \]
\[ t = \frac{-10.7 \pm \sqrt{45.76}}{0.98} \]
\[ t = \frac{-10.7 \pm 6.77}{0.98} \]
\[ t = -4.09 \, \text{or} \, 2.11 \, \text{s} \]
As time cannot be negative, the time taken for the paper to reach the balcony is 2.11 s.
Therefore, the time it takes for the paper to reach the balcony as it descends is 2.11 s.