Question

A newspaper delivery boy throws a newspaper onto a balcony 1.25 m above the

Answer 1

Answer:

(a) 3.22 m

(b) The vertical velocity, $v_y$, 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

$v_y$ = 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

$v_y$ = Vertical velocity of the paper

At maximum height, $v_y$ = 0, therefore;

$v_y$ = 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, $v_y$, 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)·$t_{tot}$

x₀ = Starting point of throwing the paper = 0

$t_{tot}$ = Total time of flight of the paper

∴ x = x₀ + v₀·cos(θ₀)·$t_{tot}$ = 0 + 15×cos(32°)×1.212 = 15.42 m

The horizontal range, of the paper = 15.42 m.