Question

A snowball is thrown off the edge of a 20 m tall building. It is thrown upward with an initial velocity in the y direction of 15 m/s and a velocity in the x direction of 5 m/s. Sketch the trajectory of the snowball from the time it is thrown until it hits the ground. Explain your trajectory 8. 9. At what angle above horizontal is the initial velocity of the snowball? 10. Atwhat speed was the snowball thrown? 11. Wnite the expressions for the and y component of the velocity of the snowball as a function of time.

Answer 1

8.

七 -t t. [大 kime

9.

$\theta=\tan ^{-1}\frac {v_y}{v_x}=\tan^{-1}\frac {15}{5}= 71.56^{\degree}$

10.

$v=\sqrt {v_x^2+v_y^2}=\sqrt {5^2+15^2}=15.81\: m/s$

11.

$v_x'=v_x$

$v_y'(t)=v_y-gt$ (For $t\leq\frac {v_y}{g}$ )

$=gt$ (For $\frac {v_y}{g}\leq t \leq\frac {2v_y}{g}$ )

$=v_y+gt$(For $t \geq\frac {2v_y}{g}$ )

12.

$t_0=\frac {v_y}{g}=\frac {15}{9.8}\: s= 1.53\: s$

$v_y'(t=2\ s)=gt=9.8\cdot (2-1.53)\:m/s= 4.60\: m/s$

$v(t=2\: s)=\sqrt {v_x^2+v_y'^2}=6.79\: m/s$

13.

$v=5\: m/s$ (At its highest point)

14.

$t_1=\frac {2v_y}{g}=\frac {2\cdot 15}{9.8}\: s= 3.06\: s$

$v_{y,hit}=\sqrt {v_y^2+2gh}=\sqrt {15^2+2\cdot 9.8\cdot 20 }\: m/s=24.84\: m/s$

$t_2=\frac {v_{y,hit}-v_y}{g}=\frac {24.84-15}{9.8}\: s= 1.00\: s$

$t=t_1+t_2=4.06\: s$

15.

$x(t)=v_xt$

$y(t)=v_yt-\frac {1}{2}gt^2$ (For $t\leq\frac {v_y}{g}$ )

$=\frac {1}{2}gt^2$ (For $\frac {v_y}{g}\leq t \leq\frac {2v_y}{g}$ )

$=v_y t+\frac {1}{2}gt^2$ (For $t \geq\frac {2v_y}{g}$ )