Question

a body is projected with initial velocity 5 I CAP + 12 J cap metre per second from origin gravity acts and negative why direction then what is the horizontal range

Answer 1

The question relates Two Dimensional Kinematics to Projectile Motion.

We will consider motion in X and Y directions individually, and go from basics of Kinematics. 


We have:

$\vec{v} = 5\hat{\imath}+12\hat{\jmath} \, \, m/s$

The two components are:

$v_x = +5 \, \, m/s \\ \\ v_y = +12 \, \, m/s$

Also, there's the gravitational acceleration acting in the negative Y direction. So,

$g = -10 \, \, m/s^2$

There is no acceleration in the horizontal direction. So, the horizontal velocity $(v_x)$ is going to remain constant.

For Range, we just need to know the Time of Flight. We can then multiply it with $v_x$ to get the answer.

Time of Flight (T) is the Time between the projection and the body hitting the ground.


So, we see when the vertical displacement becomes zero after the projection. That will give us Time of Flight.


In Y-direction,



$s=ut+\frac{1}{2}at^2 \\ \\ \\ \implies y = v_yt + \frac{1}{2}gt^2 \\ \\ \\ \implies 0 = 12t + \frac{1}{2}(-10)t^2 \\ \\ \\ \implies 5t^2-12t=0 \\ \\ \\ \implies t(5t-12)=0 \\ \\ \\ \implies t=0 \quad OR \quad t = \frac{12}{5} \\ \\ \\ \text{At t=0, body is projected.} \\ \\ \text{We need the time when it hits ground again.} \\ \\ \\ \implies T = \frac{12}{5} \, \, s$

Now, we consider the motion in X-direction. The distance travelled by the body during its Time of Flight is called Range.


In X-direction,

$v = \frac{s}{t} \\ \\ \\ \implies v_x = \frac{R}{T} \\ \\ \\ \implies R = v_x \times T \\ \\ \\ \implies R = 5 \times \frac{12}{5} \\ \\ \\ \implies \boxed{\bold{R=12 \, \, m}}$




Thus, The Range is 12 metres.

Answer 2

the answer for this question is 12 meters