Question

 A heartbroken fellow throws his engagement ring from the roof of a 20.0-m-tall building at an angle of  above the horizontal and with an initial speed of . Ignore air resistance and calculate:

a.     the maximum height above the roof of the building that the ring reaches;

b.    the speed of the ring before it reaches the ground;

c.     the horizontal range from the base of the building to the point where the ring strikes the ground.

Answer 1

Let angle $\alpha=30\degree$ , initial speed $v_0=30$ m/s

a.

$a_y=\frac{dv_y}{dt}=-g$

$v_y=-gt+v_0sin\alpha$

$h=-gt^2/2+v_0tsin\alpha$

$h'(t)=-gt+v_0sin\alpha=0$

$t=v_0sin\alpha /g=30\cdot 0.5/9.8=1.53$ s

$h_{max}=30\cdot 1.53 \cdot 0.5-9.8\cdot 1.53^2/2=11.5$ m

b.

$h=-gt^2/2+v_0tsin\alpha+h_0=0$

$h_0=20$ m

$-4.9t^2+15t+20=0$

$t=\frac{15+\sqrt{225+80\cdot 4.9}}{9.8}=4.07$ s

$v_y=-9.8\cdot 4.07+15=-24.89$ m/s

$v_x=v_0cos\alpha=30\sqrt{3}/2=25.98$ m/s

$v=\sqrt{v_x^2+v_y^2}=\sqrt{24.89^2+25.98^2}=35.98$ m/s

c.

$s=v_xt=v_0tcos\alpha=25.98\cdot 4.07=105.74$ m