Question

A stone is thrown upward, from ground level, with an initial velocity of 10.0 m/s.

a) What is the velocity of the stone after 0.50 s?
b) How high above ground level is the stone after 0.50 s?

Answer 1

The velocity of the stone after 0.50 seconds is 5.095 m/s upwards, and the height of the stone above ground level after 0.50 seconds is 2.55 meters.

Step-by-step explanation:

A stone is thrown upward from the ground with an initial velocity of 10.0 m/s. To solve for the velocity of the stone after 0.50 s and its height above ground level after 0.50 s, we can use the equations of motion under constant acceleration due to gravity, which is 9.81 m/s2 downwards.

Part a: Velocity after 0.50 s

We can use the equation:

v = u + at

Where:

v is the final velocity

u is the initial velocity (10.0 m/s upwards)

a is the acceleration due to gravity (-9.81 m/s2, negative because it is in the opposite direction of the initial velocity)

t is the time (0.50 s)

This gives us:

v = 10.0 m/s - (9.81 m/s2 × 0.50 s)

v = 5.095 m/s upwards

Part b: Height after 0.50 s

Then we can use the equation:

s = ut + (1/2)at2

Where:

s is the displacement (height above ground)

Other variables as previously defined

This gives us:

s = 10.0 m/s × 0.50 s + (1/2) × (-9.81 m/s2) × (0.50 s)2

s = 2.55 m above ground level