A stone is thrown with an initial upward velocity of 7.0 m/s and experiences negligible air resistance. If we take upward as the positive direction, what is the velocity of the stone after 0.50 s?

2.1 m/s

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 7 m/s

Time (t) = 0.50 s

Final velocity (v) =?

Since the is thrown upward, it means that the stone is going against gravity. Thus, the value of the acceleration due to gravity will be negative i.e – 9.8 m/s²

Thus, we can obtain the velocity of the stone after 0.50 s as follow:

Initial velocity (u) = 7 m/s

Time (t) = 0.50 s

Acceleration due to gravity (g) = – 9.8 m/s²

Final velocity (v) =?

v = u + gt

v = 7 + (– 9.8 × 0.5)

v = 7 – 4.9

v = 2.1 m/s

Therefore the velocity of the stone after 0.5 s is 2.1 m/s.

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

uio

explanation: