Calculating the Velocity of a Stone Thrown Up

This problem asks for the velocity of a stone thrown up vertically after a specific time, given its total time of flight. The motion of the stone is influenced only by gravity, which acts downwards. This is a classic example of vertical motion under constant acceleration (acceleration due to gravity).

Understanding Time of Flight for a Stone Thrown Up

The total time the stone thrown up stays in the air before being caught by the thrower is called the time of flight. When an object is thrown vertically upwards and returns to the same level, the time taken to reach the highest point (time of ascent) is equal to the time taken to fall back from the highest point (time of descent).

Given:

• Total time of flight = 6 seconds

Therefore, the time taken by the stone to reach its highest point is:

Time of ascent = &(\text{Total time of flight}) / 2&

Time of ascent = &6 \text{ sec} / 2 = 3 \text{ sec}&

Finding the Initial Velocity

At the highest point of its trajectory, the instantaneous velocity of the stone thrown up is zero. We can use the first equation of motion under constant acceleration to find the initial velocity (u) with which the stone was thrown. The equation is:

&v = u + at&

For upward motion, acceleration (a) is negative due to gravity, so &a = -g&. The value of acceleration due to gravity (g) is given as 9.8 m/s². At the highest point:

• Final velocity (&v&) = 0 m/s
• Time taken (&t&) = Time of ascent = 3 seconds
• Acceleration due to gravity (&g&) = 9.8 m/s²

Substituting these values into the equation &v = u - gt&:

&0 = u - (9.8 \text{ m/s}^2 \times 3 \text{ s})&

&0 = u - 29.4 \text{ m/s}&

&u = 29.4 \text{ m/s}&

So, the initial velocity of the stone thrown up was 29.4 m/s.

Calculating Velocity After 2 Seconds

Now we need to find the velocity of the stone after 2 seconds from the time of throwing. We can use the same equation of motion, &v = u - gt&, but this time with &t = 2& seconds and the calculated initial velocity &u = 29.4& m/s.

• Initial velocity (&u&) = 29.4 m/s
• Time (&t&) = 2 seconds
• Acceleration due to gravity (&g&) = 9.8 m/s²

Substituting these values:

&v = 29.4 \text{ m/s} - (9.8 \text{ m/s}^2 \times 2 \text{ s})&

&v = 29.4 \text{ m/s} - 19.6 \text{ m/s}&

&v = 9.8 \text{ m/s}&

The velocity of the stone after 2 seconds of throwing is 9.8 m/s upwards.

This matches option 1.