Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

(value of g= the acceleration due to gravity. we assume no air resisitance in ea

ID: 1430608 • Letter: #

Question

(value of g= the acceleration due to gravity. we assume no air resisitance in each example)

a) On a planet #1, you drop a stone from rest, 80 m above the ground, and the stone hits the ground 3.3 seconds later. what is the value of g for planet #1 (answer in m/s^2)

b)On planet #2, you launch a projectile straight up from the ground at a speed of 30 m/s. The projectile reaches a maximum height of 984.0 m before falling back to the ground. What is the value og g for planet #2? (answer in m/s^2)

c) On planet #3, you fire a projectile horixontally from the edge of a vertical cliff 80 m above the ground, with an initial speed of 43.0 m/s. THe projectile lands 683.0 m away from the base of the cliff. What is the value of g for planet #3? (answer in m/s^2)

d) On planet #4, You fire a projectile from the ground at an initial speed of 40 m/s at an angle of 30 degrees above the horizontal. The projectile lands 102.0 m away from the launch point. what is the value of g for planet #4? (answer in m/s^2)

e) On plane #5, a stone dropeed from a height of 200 m above the ground will hit the ground at a speed of 37.0 m/s. What is the value of g for planet #5

a) h0 l 60 S984.co atvelaaity c) 53.O 43.0 213.on 2 1) 102.0m (10 n 102.0m(40%)" SinCO (-0.30) 02.09 1600, o smoo g=-4.78 2

Explanation / Answer

everything looks fine to me

d)
Here you need to use formula for range of projectile
Range = Vo^2 * sin (2*thetha) /g
102 = (40)^2 * ( sin (2*30) ) / g
102 = 40^2 * 0.866 / g
g = 13.6 m/s^2
Answer: 13.6 m/s^2