Answer:
36.87°
Explanation:
The frog is making a projectile motion, so the time of flight for each jump is calculated as
[tex]t=\frac{2v_0\sin\theta}{g}[/tex]Where vo is the initial velocity, g is the gravity and θ is the angle. Solving for sinθ, we get
[tex]\begin{gathered} tg=2v_0\sin\theta \\ \sin\theta=\frac{tg}{2v_0} \end{gathered}[/tex]Replacing t = 0.60 s, v0 = 5.0 m/s, and g = 10 m/s², we get
[tex]\begin{gathered} \sin\theta=\frac{(0.60\text{ s\rparen}(10\text{ m/s}^2\text{\rparen}}{2(5.0\text{ m/s\rparen }} \\ \sin\theta=0.6 \\ \theta=\sin^{-1}(0.6) \\ \theta=36.87 \end{gathered}[/tex]Therefore, the angle is 36.87°
