Answer:
4.78 m/s²
Explanation:
First, we need to calculate the gravity of the planet, so we need to divide the weight by the mass to get:
g = W/m = 180N/55 kg = 3.27 m/s
Then, the range of the leap and the initial speed are related with the following equation
[tex]R=\frac{v_0^2sin(2\theta)}{g}[/tex]Where θ is the initial angle, g is the gravity and V0 is the initial velocity. Solving for V0, we get:
[tex]\begin{gathered} Rg=v_0^2sin(2\theta) \\ \frac{Rg}{sin(2\theta)}=v_0^2 \\ v_0=\sqrt{\frac{Rg}{sin(2\theta)}} \end{gathered}[/tex]Finally, we can replace R = 3.5 m, g = 3.27 m/s², and θ = 15° to get:
[tex]\begin{gathered} v_0=\sqrt{\frac{(3.5m)(3.27\text{ m/s}^2)}{sin(2(15))}} \\ v_0=\sqrt{\frac{(3.5m)(3.27\text{ m/s}^2)}{sin(30)}} \\ v_0=4.78\text{ m/s} \end{gathered}[/tex]Therefore, the answer is 4.78 m/s²
