Question

A ball is thrown straight up with a velocity of 19 m/s; what will be its velocity 3 s after being released? a) 4.9 m/s b) 9.8 m/s c) 14.7 m/s d) -19 m/s

Answer 1

The velocity of the ball 3 seconds after being released will be -19 m/s. Thus the correct option is (d).

Step-by-step explanation:

When a ball is thrown straight up, its velocity changes due to the acceleration caused by gravity. In this case, the initial velocity (u) is 19 m/s, and we need to find the final velocity (v) after 3 seconds. The formula relating initial velocity, final velocity, time, and acceleration is given by:

`\[v = u + (gt)\]`

where:

- v is the final velocity,

- u is the initial velocity,

- g is the acceleration due to gravity (approximately 9.8 m/s²),

- t is the time.

Thus the correct option is (d).

Substituting the given values into the formula:

`\[v = 19 - (9.8 * 3)\]\[v = 19 - 29.4\]\[v = -10.4\]`

Therefore, the final velocity after 3 seconds is -10.4 m/s. The negative sign indicates that the velocity is directed downward, which is consistent with the ball moving against the gravitational force.

In the context of the provided options, the closest value to -10.4 m/s is (d) -19 m/s. It's important to note that the negative sign represents the direction of the velocity rather than the magnitude. Thus, option (d) is the correct choice for the velocity of the ball 3 seconds after being released.