Answer to Question #185355 in Mechanics | Relativity for Cool

Question

A man stands on the roof of a 15.0-m-tall building and throws a rock with a velocity of magnitude 30 m/s at an angle of 33° above the horizontal. You can ignore air resistance. Calculate. A) the maximum height above the roof reached by the rock; B) the magnitude of the velocity of the rock just before it strikes the ground; and C) the horizontal range from the base of the building to the point where the rock strikes the ground.

Answer 1

(a) At maximum height, i.e. at point P:

Vertical component of velocity $v_y =0$

Initial horizontal speed

$x_{0x}=v_0cosθ \\ = 30 \times cos 33º \\ = 25.2 \;m/s$

Initial vertical speed

$v_{0y}=v_0sinθ \\ = 30 \times sin 33º \\ = 16.3 \;m/s$

Let H be the maximum height reached by the rock, as measured from the roof.

$v^2_y=v^2_{0y}-2gH$

Maximum height

$H= \frac{v^2_{0y}}{2g} \\ = \frac{16.3}{2 \times 9.8} \\ = 13.6 \;m$

(b) Magnitude of velocity of the rock just before it strikes the ground

$v = \sqrt{v^2_x+v^2_y} \\ v_x = v_{0x} \\ = 25.2 \;m/s$

Time of flight (t) is given by

$-h = v_{0y}t- \frac{1}{2}gt^2 \\ 4.9t^2 -16.3t -15 =0 \\ t = \frac{16.3± \sqrt{16.3^2 + 4 \times 4.9 \times 15}}{9.8} \\ = \frac{16.3±23.7}{9.8}$

= 4.08 s (ignoring the negative value)

$v_y = 16.3 – 9.8 \times 4.08 \\ = -23.7 \;m/s \\ v = \sqrt{v^2_x + v^2_y} \\ = \sqrt{25.2^2 + (-23.7)^2} \\ = 34.6 \;m/s$

(c) Required horizontal distance $(x) = v_{0x}t$

$= 25.2 \times 4.08 \\ = 102.8 \;m$