Question

An object that is launched from the top of the tower is dense enough that air resistance can be ignored. The object's initial velocity has components (vx = 30 m/s, vy = 20 m/s). What are the components of the object's velocity 3 seconds after launch?

Answer 1

Answer:

The components of the object's velocity 3 seconds after launch are

$v_x = 30$ m/s and $v_y=-9.43$ m/s.

Explanation:

The given components of initial velocity are

$v_x= 30$ m/s,

$v_y= 20$ m/s.

Assuming the y-axis is along the direction of gravitational force, so, the velocity along the x-axis will remain unchanged.

So, the x-componebt of the velocity after 3 seconds, $v_x = 30$ m/s.

Using the equation of motion along the y-direction, the final velocity, v, of a mass after time, t, having initial velocity, u is

v=u+at

In this case, $u=y_y= 20 m/s, a=-9.81 m/s^2$ ( as upward direction is taken as positive)

So, at time, t=3s, the final velocity in y-direction is

$v= 20+(-9.81)\times 3$

$\Rightarrow v=20-29.43$

$\Rightarrow v= -9.43 m/s$

So, the y-component of the final velocity is 9.43 m/s in the downward direction.

Hence, the components of the object's velocity 3 seconds after launch are

$v_x = 30 m/s$ and $v_y=-9.43 m/s$.

Answer 2

The final vertical velocity of the object after 3 seconds is 49.4 m/s.

The final horizontal velocity of the object is 30 m/s.

The given parameters;

• initial horizontal velocity, Vx = 30 m/s
• initial vertical velocity, Vy = 20 m/s
• time of motion, t = 3 s

The final vertical velocity of the object after 3 seconds is calculated as follows;

$v_f_y = v_0_y + gt\\\\v_f_y = 20 + 3(9.8)\\\\v_f_y = 49.4 \ m/s$

The final horizontal velocity of the object is the same as the initial horizontal velocity because horizontal motion is not affected by gravity.

$v_f_x = v_0_x = 30 \ m/s$

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