On planet number 3 you fire a projectile horizontally from the edge of a vertical cliff 80m above the ground with an initial speed of 60.2 m/s. The projectile lands 5.81 m away from the base of the cliff. What is the value of g for the planet number 3
Solution) vertical cliff height Y = 80 m
Initial horizontal velocity Vx = 60.2 m/s
Horizontal distance X = 5.81 m
g value for planet number 3 = ?
We have
Y = Vy(T) + (1/2)gT^2
Here Vy is initial vertical velocity and Vy = 0 as there is no initial vertical velocity since projectile is fired in horizontal direction.
T is time
g is acceleration of planet number 3
- 80 = 0 + (1/2)(-g)T^2
T^2 = (80×2)/g
T = (160/g)^(1/2)
We have
Horizontal displacement
X = Vx(T) + (1/2)ax(T^2)
Here ax is horizontal acceleration =0
X = Vx(T)
X = Vx(160/g)^(1/2)
5.81 = 60.2(160/g)^(1/2)
(g/160)^(1/2) = (60.2/5.81)
Squaring on both sides
(g/160) = (60.2/5.81)^2
(g/160) = 107.35
g = 160 × 107.35 = 17176 m/s^2
On planet number 3 you fire a projectile horizontally from the edge of a vertical cliff...
On planet number 3 you fire a projectile horizontally from the edge of a vertical cliff 80m above the ground with an initial speed of 60.2 m/s. The projectile lands 5.81 m away from the base of the cliff. What is the value of g for the planet number 3
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