Experiments on various alien planets, each with its own different g value (accel

Question

Experiments on various alien planets, each with its own different g value (acceleration due to gravity). We assume no air resistance for each example.

a) On a planet #1, you drop a stone from rest, 80 m above the ground, and the stone hits the ground 9.0 seconds later. What is the value of g for planet #1 (answer in m/s2)

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 261.0 m before falling back to the ground. What is the value of g for planet #2? (answer in m/s2)

c) On planet #3, you fire a projectile horixontally from the edge of a vertical cliff 80m above the ground, with an initial speed of 73.0 m/s. The projectile lands 786.0 m away from the base of the cliff. What is the value of g for planet #3? (answer in m/s2)

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 152.0 m away from the launch point. what is the value of g for planet #4? (answer in m/s2)

e) On plane #5, a stone dropeed from a height of 200 m above the ground will hit the ground at a speed of 63.0 m/s. What is the value of g for planet #5 (answer in m/s2)

Could you please include the formulas for each. Thank you!

Explanation / Answer

a)

Vi = initial velocity = 0 m/s

Y = vertical displacement = 80 m

t = time taken = 9 sec

a = acceleration

Using the equation

Y = Vi t + (0.5) a t2

80 = 0 (9) + (0.5) a (9)2

a = 1.98 m/s2

so g = 1.98 m/s2

b)

consider bottom to Top motion

Vi = initial velocity = 30 m/s

Y = vertical displacement = 261 m

Vf = final velocity at the Top = 0 m/s

a = acceleration

using the equation

Vf2 = Vi2 + 2 a Y

02 = 302 + 2 a (261)

a = - 1.72

so g = 1.72 m/s2

c)

Along the horizontal direction :

V = speed = 73 m/s

X = distance travelled = 786 m

t = time taken = X/V = 786 / 73 = 10.8 sec

Along Y-direction :

Vi = initial velocity = 0 m/s

Y = vertical displacement = 80 m

t = time taken = 10.8 sec

a = acceleration

Using the equation

Y = Vi t + (0.5) a t2

80 = 0 (10.8) + (0.5) a (10.8)2

a = 1.37 m/s2

so g = 1.37 m/s2

d)

Along the horizontal direction :

V = speed = 40 Cos30 = 34.6 m/s

X = distance travelled = 152 m

t = time taken = X/V = 152 / 34.6= 4.4 sec

Along Y-direction :

Vi = initial velocity = 40 Sin30 = 20 m/s

Y = vertical displacement = 0 m

t = time taken = 4.4 sec

a = acceleration

Using the equation

Y = Vi t + (0.5) a t2

0 = 20 (4.4) + (0.5) a (4.4)2

a = - 9.1 m/s2

so g = 9.1 m/s2

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