Question

An object is projected with velocity v0 15i+20j

Answer 1

Hi. Your Question is incomplete. Correct question will be,

An object is projected with velocity v= 15i+20j considering x along horizontal axis and y along vertical axis.find its velocity after 2s

Given conditions ⇒

Velocity of the Object = $15 \hat{i} + 20 \hat{j}$

Velocity along horizontal = $15 \hat{i}$

Velocity along vertical = $20 \hat{j}$

We know that the Velocity of the Object remains same in horizontal direction since no acceleration acts in horizontal in projectile motion.

∴ Velocity of the object along horizontal will remains same, i.e., $15 \hat{i}$

Velocity of the body along the vertical will change.

∴ u = 20 m/s.

Time(t) = 2 seconds.

Final velocity(v) = ?

Acceleration due to gravity = -g

= - 9.8 m/s²

Now, Using the First equation of motion,

v - u = at

⇒ v - 20 = (-9.8)2

∴ v = 20 - 19.6

⇒ v = 20 - 19.6

∴ v = 0.4 m/s. in y - direction.

∴ $v_y = 0.4$

In Vector Form,

Along Vertical direction, $v_y = 0.4\hat{j}$

Along horizontal direction $v_x = 15\hat{i}$

∴ Velocity of the Object at 2 seconds = $15\hat{i} + 0.4\hat{j}$

Magnitude of the Velocity = $\sqrt{(15)^2 + (0.4)^2}$

= $\sqrt{(225 + 0.16)}$

= $\sqrt{(225.16)}$

= 15.01 m/s.

Hence, the velocity of the object after 2 seconds is 15.01 seconds.

Hope it helps.

Answer 2

Answer:

above I sent u the answer

Explanation:

It will work out byeee