Question

The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by y=($8t - 5{t^2}$) m and x=6t meter, where t is in seconds. The velocity with which the projectile is projected at t = 0 is:
A. 8 m/sec
B. 6 m/sec
C. 10 m/sec
D. Not obtainable from the data

Answer 1

Hint: In order to denote the position of an object we define a coordinate system and an origin as a reference point. If a body moves from one position to another position then subtracting the initial position vector from the final position vector gives us displacement. Rate of change of displacement with respect to time gives us velocity.
Formula used:
$\dfrac{{dx}}{{dt}} = {v_x}$
$\dfrac{{dy}}{{dt}} = {v_y}$

Complete answer:
Rate of change of displacement with respect to time gives us the velocity and the rate of change of velocity with respect to time gives us the acceleration. Now if we are given with acceleration as a function of time and we were asked to find out the displacement then first we should get the velocity function by integrating the acceleration with in the proper limits given and then later after getting the velocity function, integrate it again with respect to time to get the displacement.
$\dfrac{{dx}}{{dt}} = {v_x}$
Where ‘${v_x}$’ is the velocity in x direction and ‘x’ is the displacement in x direction and ‘t’ is the time.
$\dfrac{{dy}}{{dt}} = {v_y}$
Where ‘${v_y}$’ is the velocity in x direction and ‘y’ is the displacement in y direction and ‘t’ is the time.
We have
y=($8t - 5{t^2}$) and x=6t
For velocity in x direction
$\eqalign{
  & \dfrac{{dx}}{{dt}} = {v_x} \cr
  & \Rightarrow \dfrac{{d\left( {6t} \right)}}{{dt}} = {v_x} \cr
  & \therefore {v_x} = 6 \cr} $
For velocity in x direction
$\eqalign{
  & \dfrac{{dy}}{{dt}} = {v_y} \cr
  & \Rightarrow \dfrac{{d\left( {8t - 5{t^2}} \right)}}{{dt}} = {v_y} \cr
  & \Rightarrow {v_y} = 8 - 10t \cr
  & at{\text{ t = 0}} \cr
  & \therefore {v_y} = 8 \cr} $
We got initial velocity both x and y directions. So the resultant velocity will be
$\eqalign{
  & v = \sqrt {{{\left( {{v_y}} \right)}^2} + {{\left( {{v_x}} \right)}^2}} \cr
  & \Rightarrow v = \sqrt {{{\left( 8 \right)}^2} + {{\left( 6 \right)}^2}} \cr
  & \therefore v = 10m/s \cr} $

Hence option C is the correct answer.

Note:
It is given that there is no surrounding atmosphere. That means there is no acceleration due to gravity. This happens when rms velocity of gases in that atmosphere is more than the escape velocity for that planet at the surface. That is the reason why it doesn't have any atmosphere on its surface.