Question

A marble launcher shoots a marble horizontally from the height of 0.2 m above a horizontal

floor. The marble lands on the floor 5 m away from the launcher. How long did the marble stay

in the air?

A. 0.1 s B. 0.2 s C. 0.3 s D. 0.4 s​

Answer 1

Given:

A marble launcher shoots a marble horizontally from the height of 0.2 m above a horizontal floor. The marble lands on the floor 5 m away from the launcher.

To find:

Time of flight?

Calculation:

Let horizontal velocity of projection be v:

$\rm \therefore \: t = \sqrt{ \dfrac{2h}{g} }$

Now, range of flight will be :

$\rm \: R = v \times t$

$\rm \implies\: R = v \times \sqrt{ \dfrac{2h}{g} }$

$\rm \implies\: 5= v \times \sqrt{ \dfrac{2 \times 0.2}{g} }$

$\rm \implies\: 5= v \times \sqrt{ \dfrac{2 \times 0.2}{10} }$

$\rm \implies\: 5= v \times \sqrt{0.2 \times 0.2 }$

$\rm \implies\: 5= v \times 0.2$

$\rm \implies\:v = 25 \: m {s}^{ - 1}$

Now, put value of v :

$\rm \: R = v \times t$

$\rm \implies\: 5= 25 \times t$

$\rm \implies\: t = \dfrac{1}{5}$

$\rm \implies\: t = 0.2 \: sec$

So, time of flight is 0.2 sec.

Answer 2

Answer:

b

Explanation:

b for ball