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 15.0 m/s and an angle of 32° above horizontal, find:
L/10]
Draw a detailed diagram. g = 9.81 m/s2 [down]
(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

(a) 3.22 m

(b) The vertical velocity, , at maximum height is 0 m/s, the horizontal velocity, vₓ, is 12.72 m/s

(c) The acceleration at maximum height = g = 9.81 m/s²

(d) The time it takes for the paper to reach the balcony is 1.212 seconds

(e) The horizontal range, of the paper is 15.42 m.

Explanation:

(a) Given that we re given a projectile motion, we have the following governing equations;

y = y₀ + v₀·sin(θ₀)·t - 0.5×g·t²

= v₀·sin(θ₀) - g·t

Where:

y = Height of the paper

y₀ = Initial height of the paper = Ground level = 0

v₀ = Inititial velocity of the paper = 15.0 m/s

θ₀ = Angle in which the paper is thrown = 32° above the horizontal

g = Acceleration due to gravity = 9.81 m/s²

t = Time taken to reach the height h

= Vertical velocity of the paper

At maximum height, = 0, therefore;

= v₀·sin(θ₀)·t - g·t = 0

v₀·sin(θ₀) = g·t

t = v₀·sin(θ₀)/g = 15×sin(32°)/9.81 = 0.81 seconds

y = y₀ + v₀·sin(θ₀)·t - 0.5×g·t² = 0 + 15×sin(32°)×0.81-0.5×9.81×0.81² = 3.22 m

(b) The vertical velocity, , at maximum height = 0 m/s, the horizontal velocity, vₓ, = 15×cos(32°) = 12.72 m/s

(c) The acceleration at maximum height = g = 9.81 m/s²

(d) The time it takes to maximum height = 0.81 seconds

The time the paper will take to fall to 1.25 m above the ground, which is 3.22 - 1.25  = 1.97 meters below maximum height is therefore given as follows;

y = y₀ + v₀·sin(θ₀)·t - 0.5×g·t²

Where:

v₀ = 0 m/s at maximum height

y = -1.97 m downward motion

y₀ = 0 starting from maximum height downwards

1.97 = 0 + 0·sin(θ₀)·t - 0.5×9.81×t²

-1.97 =  - 0.5×9.81×t²

t = (-1.97)/(-0.5*9.81) = 0.402 seconds

The time the paper will take to fall to 1.25 m above the ground = 0.81+0.402 = 1.212 seconds

Therefore, the time it takes for the paper to reach the balcony = 1.212 seconds

(e) The horizontal range, x, is given by the relation;

x = x₀ + v₀·cos(θ₀)·

x₀ = Starting point of throwing the paper = 0

= Total time of flight of the paper

∴ x = x₀ + v₀·cos(θ₀)· = 0 + 15×cos(32°)×1.212 = 15.42 m

The horizontal range, of the paper = 15.42 m.

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.

he made 11 calls in one day