Question

A body is thrown with a velocity of 50 m/s at an angle of 37° with the horizontal. Find Direction of the velocity after 4 sec [take g = 10 m/s²]

✅The que seems to be too short but it takes lot of time to solve this. Let's see who will solve.​

Answer 1

Topic :-

Kinematics

Given :-

A body is thrown with a velocity of 50 m/s at an angle of 37° with the horizontal.

To Find :-

Direction of the velocity of body after 4 seconds.

Solution :-

Projectile's velocity, u = 50 m/s

Projectile's angle, θ = 37°

g = 10 m/s²

We will be analysing projectile motion in horizontal and vertical motion.

Horizontal Motion (along X-axis),

The horizontal component of projectile's velocity remains unchanged because there is no acceleration in the horizontal direction.

$\sf{Horizontal\:component\:of\:projectile's\:velocity,\overrightarrow{v_x} = (u\cdot cos\theta)\:\hat{i}}$

$\sf{Horizontal\:component\:of\:projectile's\:velocity,\overrightarrow{v_x} = 50\cdot\dfrac{4}{5}\:\hat{i}}$

$\sf{\left( \because \cos 37^{\circ}=\dfrac{4}{5}\right)}$

$\sf{Horizontal\:component\:of\:projectile's\:velocity,\overrightarrow{v_x} = 40\:\hat{i}}$

Calculating Time of Flight,

$\sf {Time\:of\:flight=\dfrac{2u\sin\theta}{g}}$

$\sf {Time\:of\:flight=\dfrac{2\times50\times3}{10\times 5}\:seconds}$

$\sf {\left(\because \sin 37^{\circ}=\dfrac{3}{5}\right)}$

$\sf {Time\:of\:flight=\dfrac{300}{50}\:seconds}$

$\sf {Time\:of\:flight=6\:seconds}$

We can conclude that body completed its flight in 6 seconds and it was at its highest point at t = 3 seconds.

Vertical Motion (along Y-axis),

The body will return to ground after reaching at its highest point.

At highest point, vertical component of projectile's velocity is zero.

So, this case gets similar to a case in which a body is dropped from some height.

So in this case,

$\sf {\overrightarrow{u'_y}=0\:\hat{j}}$

$\sf{\overrightarrow{a}=-10 \:\hat{j}}$

$\sf{t'=4\:s-3\:s=1\:s}$

Applying equation of motion,

$\sf{\overrightarrow{v_y}=\overrightarrow{u'_y}+\overrightarrow{a_y}\cdot t'}$

$\sf {\overrightarrow{v_y}=0\:\hat{j}-10\:\hat{j}\cdot (1)}$

$\sf {\overrightarrow{v_y}=-10\:\hat{j}}$

$\sf {Velocity\:in\:vertical\:direction,\overrightarrow{v_y}=-10\:\hat{j}}$

Calculating direction of velocity,

$\sf{\overrightarrow{V}=\overrightarrow{v_x}+\overrightarrow{v_y}}$

$\sf{\overrightarrow{V}=40\:\hat{i}-10\:\hat{j}}$

Answer :-

Direction of velocity after 4 seconds :-

$\underline{\boxed{\sf{\overrightarrow{V}=40\:\hat{i}-10\:\hat{j}}}}$

Note :

Direction of velocity in vertical could have been easily calculated using :-

$\sf{\overrightarrow{v_y}=\overrightarrow{u_y}+\overrightarrow{a_y}\cdot t}$

$\sf{\overrightarrow{v_y}=(u\sin\theta-g\cdot t)\:\hat{j}}$

$\sf{\overrightarrow{v_y}=\left(50\times \dfrac{3}{5}-10(4)\right)\:\hat{j}}$

$\sf{\overrightarrow{v_y}=\left(30-40\right)\:\hat{j}}$

$\sf{\overrightarrow{v_y}=-10\:\hat{j}}$

Answer 2

Question:

A body is thrown with a velocity of 50 m/s at an angle of 37° with the horizontal. Find Direction of the velocity after 4 sec [take g = 10 m/s²].

Solution:

We have initial velocity = 50m/s

Break it into horizontal and vertical component i.e ucos37° and usin37°

Ucos 37° = $50\times\frac{4}{5}$ => $40\hat{i}$

Usin 37° = $50\times\frac{3}{5}$ => $30\hat{j}$

Hence, initial velocity vector is $\bf 40 \hat{i} + 30 \hat{j}$

Since, this is a projectile motion, the acceleration vector will be directed vertically along the gravity.

That's a= - 10 $\hat{j}$

Using equation of motion, v=u+at

We have to find the velocity vector at t=4s

Therefore,

$v = (40 \hat{i} + 30 \hat{j}) +(-10 \hat{j}). 4 \\\\ v= (40 \hat{i} + 30 \hat{j}) +(-40 \hat{j}) \\\\ \implies (40 \hat{i} + 30 \hat{j}-40 \hat{j}) \\\\ \implies 40 \hat{i} - 10 \hat{j}$

This was simple, hardly took 3 minutes.