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...

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....

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.