Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
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
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
