Answer:
Approximately [tex]1.90\; {\rm m}[/tex] (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] and that air resistance on the bullfrog is negligible.)
Explanation:
It is given that the initial velocity [tex]u[/tex] is at an angle of [tex]\theta = 37.0^{\circ}}[/tex] above the horizontal. Therefore:
If air resistance on the bullfrog is negligible, final vertical velocity right before landing will be the opposite of the initial vertical velocity:
[tex]v_{y} = (-u_{y}) = (-2.64799)\; {\rm m\cdot s^{-1}}[/tex].
Change in vertical velocity:
[tex](v_{y} - u_{y}) = (-2.64799) \; {\rm m\cdot s^{-1}} - 2.64799 \; {\rm m\cdot s^{-1}} \approx (-5.29598)\; {\rm m\cdot s^{-1}}[/tex].
Vertical acceleration of the bullfrog will be [tex]a = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex] while the frog is in the air. Therefore, time required for the velocity to change will be:
[tex]\begin{aligned} t &= \frac{v_{y} - u_{y}}{a_{y}} \\ &= \frac{v_{y} - u_{y}}{(-g)} \\ &\approx \frac{(-5.29598)\; {\rm m\cdot s^{-1}}}{(-9.81)\; {\rm m\cdot s^{-2}}} \\ &\approx 0.53985\; {\rm s}\end{aligned}[/tex].
If air resistance on the bullfrog is negligible, horizontal velocity will be constant: [tex]v_{x} = u_{x} \approx 3.51400\; {\rm m\cdot s^{-2}}[/tex]. After [tex]t \approx 0.53985\; {\rm s}[/tex] in the air, the horizontal displacement of the bullfrog will be:
[tex]\begin{aligned}x_{x} &= v_{x}\, t \\ &\approx 3.51400\; {\rm m\cdot s^{-1}} \times 0.53985\; {\rm s} \\ &\approx 1.90\; {\rm m} \end{aligned}[/tex].
Therefore, this bullfrog will cover a distance of approximately [tex]1.90\; {\rm m}[/tex].
