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.

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.