Question

The velocity of a particle at t=0 is u=4i+4j and its acceleration is -10j.At t=0 the particle is at 1,0 m

Answer 1

Answer:

The x coordinate of the point where its y coordinate is again zero is 3.2.

Explanation:

The complete question is:

The velocity of a particle at t = 0 is  u = (4i+4j) m/s and its acceleration is  -10 j m/s^2. At t = 0, the particle is at  (1, 0) m. The x - coordinate of the point,  where its y-coordinate is again zero is​:

Solution:

velocity of the particle

$\vec u=4\hat i+4\hat j$

i.e. velocity of the particle in y direction is 4 m/s

Acceleration of the particle

$\vec a=-10\hat j$

i.e. acceleration in the y-direction is -10 m/s²

At t = 0, the y-displacement is zero

Let after time t its displacement is again zero then

Using the second equation of motion in the y-direction

$\boxed{h=ut+\frac{1}{2}at^2}$

$0=4\times t+\frac{1}{2}\times (-10)\times t^2$

$\implies 0=4t-5t^2$

$\implies t(4-5t)=0$

$\implies t=0, \frac{4}{5}$

$\implies t=\frac{4}{5}$ sec

The velocity of the particle in y-direction is 4 m/s

Since there is no acceleration in x-direction

Therefore, the displacement in the x-direction

$=\text{speed}\times \text{time}$

$=4\times\frac{4}{5}}$

$=3.2$ m

Thus the x coordinate of the point where its y coordinate is again zero is 3.2.

Hope this helps.