Question

Please helpppp... thank u ❤️​

Answer 1

Answer:

For Ryan:

$at \: highest \: point : v = 0 \\ velocity = 0 \: {ms}^{ - 1}$

For the football kicked:

$s = ut - \frac{1}{2} g {t}^{2} \\ for \: range : s = 0 \\ u \sin( \theta) = \frac{gt}{2} \\ t = \frac{2u \sin(\theta) }{g} \\ range(x) = ut \\ x = \frac{ {2u {}^{2} \sin(\theta) }^{} }{g} \\ \frac{50 \times 9.8}{ {2318.9}^{} } = \sin(\theta) \\ \\but \: 0 = u \sin(25) - 2 \times 9.8 \times 50 \\ u = 2318.9 \: {ms}^{ - 1} \\ \theta = 12.2 \degree$

Answer 2

Answer:

the highest point for the first question is 0 meters per second

the answer for question 2 is 45 degrees