A newspaper delivery boy throws a newspaper onto a balcony 1.25 m above the height of his hand when he releases the paper. Given that he throws the paper with a speed of 25.0 m/s and an angle of 32° above horizontal, find:
(a) the maximum height of the paper’s trajectory
(b) the velocity at maximum height
(c) the acceleration at maximum height
(d) the time it takes for the paper to reach the balcony
(e) the horizontal range of the paper
Let's solve this problem step by step using kinematic equations:
(a) To find the maximum height of the paper's trajectory, we need to use the vertical component of its initial velocity, which is given by V₀sin(θ) = 25.0 m/s × sin(32°) = 13.4 m/s. The maximum height can be calculated using the formula:
Δy = (V₀sin(θ))² / 2g
where g is the acceleration due to gravity, approximately 9.81 m/s². Thus, Δy = (13.4 m/s)² / (2 × 9.81 m/s²) = 9.19 m. Therefore, the maximum height of the paper's trajectory is 9.19 m above its initial position.
(b) At the maximum height, the vertical component of the paper's velocity is zero, but its horizontal component remains unchanged at V₀cos(θ) = 25.0 m/s × cos(32°) = 21.2 m/s. Therefore, the velocity at maximum height is V = 21.2 m/s in the horizontal direction and 0 m/s in the vertical direction.
(c) At the maximum height, the acceleration of the paper is purely vertical and equal to the acceleration due to gravity, g = 9.81 m/s².
(d) To find the time it takes for the paper to reach the balcony, we need to use the formula for the time of flight:
t = 2V₀sin(θ) / g
Thus, t = 2 × 25.0 m/s × sin(32°) / 9.81 m/s² = 4.20 s. Therefore, it takes the paper 4.20 s to reach the balcony.
(e) Finally, to find the horizontal range of the paper, we use the formula:
R = V₀cos(θ) × t
Thus, R = 25.0 m/s × cos(32°) × 4.20 s = 87.8 m. Therefore, the horizontal range of the paper is 87.8 m.