Question

A ball is projected in air with velocity v = ( 40î + 40j ) m/s . where i and j are the unit vector along horizontal and vertical axis . The horizontal distance it covers in 4s after the Start of motion is ?
options
1. 160m
2. 30
3. 20
4. 10
if you don't know the answer then don't answer this please hurry up​

Answer 1

Answer: should be 160m

Explanation:

Answer 2

Given:

A ball is projected in air with velocity v = ( 40î + 40j ) m/s .

To find:

Horizontal distance travelled in 4 seconds?

Calculation:

• We know that : Along the X axis, a projectile doesn't experience any acceleration and hence, the horizontal component of velocity remains constant.

• However, acceleration is experienced along Y axis equal to 9.8 m/s². However this acceleration is of no use in this question.

Distance along X axis :

$d_{ x} = v_{x} \times t$

$\implies \: d_{x} = 40 \times 4$

$\implies \: d_{x} = 160 \: metres$

So, distance travelled along X axis is 160 metres.