Alex J.

asked • 12/02/21# Tricky projectile physics problem

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

## 1 Expert Answer

Grigoriy S. answered • 12/02/21

AP Physics / Math Expert Teacher With 40 Years of Proven Success

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 **v**_{0} 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 **v**_{0} in y direction

g_{y} – component of vector** g** in y direction

We see that

v_{0x }= v_{0}∙cos θ g_{x} = 0

v_{0y} = v_{0}∙sin θ g_{y} = - g

Finally

v_{x} = v_{0}∙cos θ (1)

v_{y} = v_{0}∙sin θ – gt (2)

Similarly for coordinates

x = v_{0}∙cos θ∙t (3)

y = v_{0}∙sin θ∙t – ½ g∙t^{2} (4)

a) For height we use v_{y}^{2} – v_{0y}^{2} = - 2gh

Height is max, when v_{y} = 0

h_{max} = v_{0y}^{2} / 2g or *h*_{max}* = v*_{0y}^{2}*∙sin*^{2}* θ / 2g*

b) Components of velocity

*v*_{x}* = v*_{0}*∙cos θ** v*_{y}* = 0*

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

^{2}.

d) Coordinate y = v_{0}∙sin θ∙t – ½ g∙t^{2}

For balcony y = h = 1.25 m.

1.25 = 12.5 sin 53° - ½ 10t^{2}

Or

1.25 = 10t – 5t^{2}

Or

4t^{2} – 8t + 1 = 0

The bigger value * t = 2.1 s *

e) Now range *R = v*_{0}** ∙cos θ∙t **you can find by yourself

Alex J.

Hi there, Sorry I just have another quick question: What is the direction equivalent to – [40° W of S]? a) [40° E of S] b) [40° W of N] c) [40° E of N] d) [50° S of W] e) [50° E of N]12/08/21

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Grigoriy S.

12/02/21