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 horizontal distance between the boy and the balcony.
We can use the horizontal component of the velocity (v_x) to find the time it takes for the paper to reach the balcony.
v_x = v cos(θ) = 25.0 m/s cos(32°) ≈ 21.3 m/s
The horizontal distance (d) between the boy and the balcony can be found using the formula:
d = v_x t
where t is the time it takes for the paper to reach the balcony. We can find t using the vertical component of the velocity (v_y) and the height of the balcony (h).
v_y = v sin(θ) = 25.0 m/s sin(32°) ≈ 13.5 m/s
Using the formula for vertical motion:
h = v_y t - 0.5 g t^2
where g is the acceleration due to gravity (≈ 9.81 m/s^2), we can solve for t:
t = (v_y ± sqrt(v_y^2 + 2gh)) / g
We take the positive root because we want the time it takes for the paper to reach the balcony from the boy's hand.
t = (13.5 ± sqrt(13.5^2 + 2(9.81)(1.25))) / 9.81 ≈ 1.1 s
Using this value for t, we can find the horizontal distance:
d = v_x t ≈ 21.3 m/s × 1.1 s ≈ 23.4 m
Therefore, the horizontal distance between the boy and the balcony is approximately 23.4 m.
(b) The height of the balcony above the ground.
We can use the same formula for vertical motion:
h = v_y t - 0.5 g t^2
with the value of t we found earlier:
h = 13.5 m/s × 1.1 s - 0.5 (9.81 m/s^2) (1.1 s)^2 ≈ 16.7 m
Therefore, the height of the balcony above the ground is approximately 16.7 m.