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 degrees above horizontal, find:  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.  

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Thank you for the reply and correction.

If you use an angle of 32 degrees, the new answers would be:

Maximum height = 8.95 m

Horizontal velocity component = 21.2 m/s

Vertical velocity component, u_y = 13.25 m/

And time, T = 0.1 s and 2.6 s

A time of...

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Thank you for the reply and correction.

If you use an angle of 32 degrees, the new answers would be:

Maximum height = 8.95 m

Horizontal velocity component = 21.2 m/s

Vertical velocity component, u_y = 13.25 m/

And time, T = 0.1 s and 2.6 s

A time of 2.6 s will correspond to a range of 55.12 m.

Hope this helps.

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This is a numerical of projectile motion. Here, the initial velocity, u = 25 m/s and `theta`  = 35 degrees.

a) Maximum height achieved can be calculated by the following equation:

`H = (u^2sin^2theta)/(2g) = (25^2 xx sin^2(35))/(2 xx 9.81) = 10.48 m`

Thus, the newspaper achieves a maximum height of 10.48 m.

b) At the point of maximum height, the y-component of velocity is 0 m/s, while the horizontal velocity stays the same. Thus, the velocity at maximum height is same as the horizontal velocity component.

The horizontal component of velocity = u cos(35) = 25 x cos(35) = 20.48 m/s.

c) In this type of motion, the only acceleration acting on the object is the acceleration due to gravity, g. And it is acting in the downward direction always. 

d) The vertical distance traveled by the newspaper is 1.25 m (height of balcony above his hand) and is given as:

y = u_y t -1/2 gt^2

where, u_y is the vertical component of velocity = u sin (35) = 14.34 m/s

substituting all the values, we get: t = 0.09 s and 2.83 s.

The two times correspond to newspaper reaching a height of 1.25 m on its way to maximum height and on its way down. 

The time of flight = 2.83 s (as the other time is too small to need the type of velocity we have here).

e) The horizontal distance traveled by the newspaper is

x = 20.48 x 2.83 = 57.96 m

Hope this helps. 

 

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