Tricky projectile physics problem

Question

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 12.5 m/s and an angle of 53° above horizontal and the paper lands on the way down find:

(a)   the maximum height of the paper’s trajectory

(b)   the vertical and horizontal 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

Answer 1

We will select x-axis to the right and y-axis vertically up.

So

                                             v_x = v_0x + g_x∙t

                                              v_y = v_0y + g_y∙t

Here

                           v_x  - component of vector v in x direction,

                            v_0x – component of initial velocity v₀ in x direction,

                            g_x – component of vector g in x direction

                           v_y  - component of vector v in y direction,

                            v_0y = component of initial velocity v₀ in y direction

                            g_y – component of vector g in y direction

We see that

                            v_0x = v₀∙cos θ          g_x = 0

                                 v_0y  = v₀∙sin θ         g_y = - g

Finally

                                              v_x = v₀∙cos θ                                        (1)

                                              v_y = v₀∙sin θ – gt                               (2)

Similarly for coordinates

                                                 x = v₀∙cos θ∙t                                     (3)

                                                 y = v₀∙sin θ∙t – ½ g∙t²                (4)  

         

a)    For height we use v_y² – v_0y² = - 2gh

Height is max, when v_y = 0

                       h_max = v_0y² / 2g               or     h_max = v_0y²∙sin² θ / 2g

b)    Components of velocity    

                                                             v_x = v₀∙cos θ         v_y = 0

                               

c)    Acceleration of free fall g, directed always down and has magnitude 9.8 m/s².

d)    Coordinate  y = v₀∙sin θ∙t – ½ g∙t²    

For balcony y = h = 1.25 m.

                                                             1.25 = 12.5 sin 53° - ½ 10t²

Or

                                                    1.25 = 10t – 5t²

Or

                                                     4t² – 8t + 1 = 0

The bigger value  t = 2.1 s    

e)    Now range   R = v₀∙cos θ∙t you can find by yourself