Question

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?

Answer 1

The velocity of the stone after 0.50 s is 2.1 m/s upwards, calculated using the kinematic equation with the initial velocity and acceleration due to gravity.

Step-by-step explanation:

The velocity of the stone after 0.50 s can be found using the kinematic equation:

v = u + at

Where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (which is -9.8 m/s² when taking upward as the positive direction), and t is the time. Substituting the values given:

v = 7.0 m/s + (-9.8 m/s²)(0.50 s)

v = 7.0 m/s - 4.9 m/s

v = 2.1 m/s

Therefore, after 0.50 s, the velocity of the stone is 2.1 m/s upwards.

Answer 2

2.1 m/s

Step-by-step 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.