Question

If a ball is thrown vertically upwards with a velocity of 40 m/s. Then what will be the magnitude of its displacement after 6 s? /take g = 10 m/s2

Answer 1

$s = ut + \frac{1}{2} g {t}^{2} \\ s = 40(6) + \frac{1}{2} \times 10 \times {6}^{2} \\ s = 240 + 5 \times 36 = 240 + 180 \\ s = 420 \: m$

Answer 2

Answer:

$\large\bold\red{60\:m}$

Explanation:

Given,

A ball is thrown vertically upward.

Velocity ,$u = 40 \:m{s}^{-1}$

acceleration, $a = -g = -10\:m{s}^{-2}$

Here,

acceleration is acting in the direction opposite to the direction of motion.

Thus,

it's taken negative.

Also,

Time taken, $t=6\: s$

Let,

the displacement be 's' metres.

Now,

it is the case of motion in straight line.

Therefore,

we have the Equation,

$\large\bold{s = ut + \frac{1}{2} a {t}^{2} }$

Putting the respective values,

we get,

$= > s = (40 \times 6) + \frac{1}{2} \times ( - 10) \times {6}^{2} \\ \\ = > s = 240 - \frac{36 \times 10}{2} \\ \\ = > s = 240 - 180 \\ \\ = > s = 60 \: m$

Hence,

60 m is the displacement after 6 seconds.