Question

Jane and Wussle are playing siatong on the field. They both agreed that the player with the farthest siatong to land will treat the loser with a snack. Jane hit her stick with an initial velocity of 12 m/s at 30° angle while Wussle launched his siatong with 15m/s at 25°

a. How far did wussle's siatong landed?

Answer 1

a) Let's first find the time that Wussle's siatong takes to reach the maximum height:

$$v_y=v_0sin\theta-gt_{rise},$$ $$0=v_0sin\theta-gt_{rise},$$ $$t_{rise}=\dfrac{v_0sin\theta}{g}.$$

Then, we can find the total time of flight of Wussle's siatong:

$$t=2t_{rise}=\dfrac{2\cdot 15\ \dfrac{m}{s}\cdot sin25^{\circ}}{9.8\ \dfrac{m}{s^2}}=1.29\ s.$$

Finally, we can find the range of Wussle's siatong from the kinematic equation:

$$x=v_0tcos\theta=15\ \dfrac{m}{s}\cdot 1.29\ s\cdot cos25^{\circ}=17.54\ m.$$

Answer:

a) $x=17.54\ m.$