Question

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

Answer 1

Given conditions ⇒

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

Velocity along horizontal = [/tex] 15 \hat{i} [/tex]

Velocity along vertical = [/tex] 20 \hat{j} [/tex]

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

Explanation:

Given the initial velocity of the body is 15i+20 j

given g=-10

velocity in y =u +at

$[tex]we \: know \: that \: the \: velocity \: in \: the \: x \: axis \\ is \: constant \: because \: no \: acceleration \: is \: produced \\ v \: in \: x = 15m {s}^{ - 1} \\ velocity \: in \: y = u + at \\ u = 20m {s}^{ - 1} \: \: \: \: a = - 10m {s}^{ - 2} \\ y = 20 + ( - 10)2 \\ y = 20 - 20 \\ y = 0</p><p></p><p>$

$v = 15i + 0j \\ |v | = \sqrt{ {15}^{2} + {0}^{2} } \\ = \sqrt{ {15}^{2} } \\ |v | = 15 m{s}^{ - 1}$

hope it helps