Question

A baseball was hit at 45 m/s at ang angle of 45 degrees above the horizontal

a.How long did it remain in the air

b.How far did it travel horizontally

Answer 1

(a) Let's first find the time that the baseball takes to reach its maximum height:

$$v_y=v_{0y}-gt,$$ $$0=v_0sin\theta-gt,$$ $$t=\dfrac{v_0sin\theta}{g}.$$

Finally, we can find the flight tme of the baseball as follows:

$$t_{flight}=2t=\dfrac{2v_0sin\theta}{g},$$ $$t_{flight}=\dfrac{2\cdot45\ \dfrac{m}{s}\cdot sin45^{\circ}}{9.8\ \dfrac{m}{s^2}}=6.49\ s.$$

(b) We can find the horizontal distance traveled by the baseball from the kinematic equation:

$$x=v_0t_{flight}cos\theta,$$ $$x=45\ \dfrac{m}{s}\cdot6.49\ s\cdot cos45^{\circ}=206.5\ m.$$