Question

A ball is thrown vertically upward with initial velocity $30m{\sec ^{ - 1}}$ what will be its position vector at time $t = 5\sec $ taking origin at the point of projection, vertical up as positive y-axis and horizontal as x-axis?
[A] $(0,25)$
[B] $(0,20)$
[C] $(0,45)$
[D] $(0,5)$

Answer 1

Hint: In order to solve this question we need to understand the motion of the ball thrown vertically upward. While throwing vertically upward only force of gravity acting downward direction hence acceleration of ball is in downward direction equal to acceleration due to gravity $9.8m{\sec ^{ - 2}}$ since velocity in upward direction is constant so acceleration in upward direction is zero. Also here we can use the equation of motion because acceleration is constant.

Complete step-by-step solution:
Since the velocity is only in upward direction so horizontal coordinate become zero
So x-coordinate of velocity is always $0$
Let the initial velocity be, $u = 30m{\sec ^{ - 1}}$
And acceleration is given by, $a = - 10m{s^{ - 2}}$ during upward motion
Let us calculate the maximum distance it can cover during its upward motion,
Let the maximum distance be, ${y_{\max }}$
We know final velocity of ball is $v = 0$
So using equation of motion we get, ${v^2} = {u^2} + 2a{y_{\max }}$
Putting values we get, $0 = ({30^2}) - (2 \times 10 \times {y_{\max }})$
${y_{\max }} = \dfrac{{900}}{{20}}$
${y_{\max }} = 45m$
Let the time taken to cover maximum distance be ${t_1}$
So, using equation of motion, $v = u + a{t_1}$
Putting values we get, $0 = (30) - (10 \times {t_1})$
${t_1} = \dfrac{{30}}{{10}}\sec $
${t_1} = 3\sec $
According to question we need to calculate the y distance at time, $t = 5\sec $
It means during $5\sec $ the ball is coming back toward earth.
So for downward motion, $a = 10m{\sec ^{ - 2}}$
And let the distance it covers during downward fall be, $s = y$ measure from highest point
We have to calculate this distance for ${t_2} = t - {t_1}$
${t_2} = (5 - 3)\sec $
${t_2} = 2\sec $
Also initial velocity would be ${u_1} = 0m{\sec ^{ - 1}}$
Remember we are considering only downward motion from maximum position
So using equation of motion $S = ut + \dfrac{1}{2}a{t^2}$ we get, $y = {u_1}{t_2} + \dfrac{1}{2}a{t_2}^2$
Putting values we get, $y = (0 \times 2) + \dfrac{1}{2}(10)(4)$
$y = 20m$
So the y- coordinate from origin is given by, ${y_1} = {y_{\max }} - y$
${y_1} = (45 - 20)m$
${y_1} = 25m$
So the correct option is [A] $(0,25)$.

Note: It should be remembered that during motion in upward or downward direction we are neglecting air resistance for simpler calculation otherwise we know air resistance force is directly proportional to speed and surface area of ball on which it is acting and due to this the acceleration of body changes and we do not get the same result.