Question

A body projected vertically upward with speed 40m/s dis travelled by body in last second of its journey?

Answer 1

$\sf\underline{ Given :-}$

• A body is projected vertically upwards with a speed of 40m/s .

$\sf\underline{ To \ find :-}$

• The distance travelled in last second of its journey .

$\sf\underline{ Answer :-}$

The ball was projected vertically upwards with a speed of 40m/s . So we can calculate the time of ascent as ,

$\sf\implies time_{ascent}= \dfrac{u}{g} \\\\\sf\implies time_{ascent}= \dfrac{40m/s}{10m/s^2} \\\\\sf\implies \underline{ time_{ascent}= 4s }$

Now here we need to find the distance travelled in the last second of its upward motion. Hence here we need to find the distance travelled in the 4th second , as ,

$\sf\implies S_n = u +\dfrac{1}{2}a (2n-1) \\\\\sf\implies S_4 = 40m/s -\dfrac{1}{2} (10m/s^2)[ 2(4) - 1] \\\\\sf\implies S_4 = 40 - 5( 8 - 1) m \\\\\sf\implies S_4 = 40 - 5(7) m \\\\\sf\implies S_4 = 40m - 35m \\\\\sf\implies \boxed{\frak {S_4 = 5m }}$

Hence the distance travelled in the last second of its journey is 5m .

Answer 2

Explanation:

4.9m

So, the correct answer is '4.9m'.