You throw a 3.00-n rock vertically into the air from ground level. you observe that when it is 15.9 m above the ground, it is traveling at 25.7 m/s upward.

a)use the work-energy theorem to find the rock's speed just as it left the ground.

b)use the work-energy theorem to find its maximum height.

a) u = 31.197 m/s

b)

Explanation:

given,

weight = 3 N

mass = 3/ 9.81 = 0.305 kg

h = 15.9 m

velocity at which it is traveling upward = 25.7 m/s

a) using work energy principle

u = 31.197 m/s

b) At maximum height

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft