**Title: Exploring Miranda, a Satellite of Uranus**Miranda, a satellite of Uranus, is illustrated in the images below. The moon can be modeled as a sphere with a radius of 242 km and a mass of \(6.68 \times 10^{19}\) kg.**Figure Analysis:**- *Figure (a):* Shows a general view of Miranda.- *Figure (b):* Provides a magnified view of a specific region.---### Questions:(a) **Find the free-fall acceleration on its surface.**Answer: \(\_\_\_\_\) m/s\(^2\)(b) **A cliff on Miranda is 5.00 km high.** - It appears on the limb at the 11 o'clock position in Figure (a) and is magnified in Figure (b).- A devotee of extreme sports runs horizontally off the top of the cliff at 5.70 m/s.- Calculate the time interval of his flight. (Ignore the difference in \(g\) between the top and base of the cliff.)Answer: \(\_\_\_\_\) s(c) **How far from the base of the vertical cliff does he strike the icy surface of Miranda?**Answer: \(\_\_\_\_\) m(d) **What is his vector impact velocity?**Answer: \(\_\_\_\_\) m/s \(\_\_\_\_\)° below the horizontal---The diagrams depict Miranda’s surface features and the cliff's location, highlighting the differences in terrain and elevation. The task involves applying physics principles to calculate gravitational effects, projectile motion, and velocity vectors at the high cliff's location on Miranda.

The radius of the satellite (R) = 242 km = 242×103 m The mass of the satellite (M) = 6.68×1019 kg The height of the cliff in Miranda (H) = 5 km =5×103 m. The horizontal velocity ux=5.70 m/s. The free-fall acceleration on the surface of Miranda needs to be determined. The time of flight has to be calculated. The horizontal distance from the base of the cliff where the devotee of extreme sports land, needs to be determined. The impact velocity and the angle has to be calculated.(a) The free-fall acceleration on the surface g=GMR2=6.67×10-11×6.68×1019242×1032=0.07608 m/s2 (b) A projectile is fired horizontally from the top of the 5km high H=5000 m cliff with horizontal velocityux=5.70 m/s, the time of flight (t) is given by, t=2Hg=2×50000.07608=362.55 s. (c) The horizontal distance from the base of the cliff to the projectile hitting location on the ground, X =uxt X =5.70×362.55=2.0665×103 mvx=ux=5.70 m/svy=uy+-gt=0 -0.07608×362.55 =-27.58 m/sThe impact velocity v=vx2+vy2=28.16 m/s tanθ=vyvx=-27.585.70θ=tan-1-27.585.70=-78.320θ=78.320 below the horizontal(a) Free fall acceleration = 0.07608 m/s2 (b) Time of flight of the projectile = 362.55 s (c) The horizontal distance from the base of the cliff = 2066.5 m (d) Impact velocity = 28.16 m/s and angle = 78.320 below the horizontal

Answered: Miranda, a satellite of Uranus, is…