Question

A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with magnitude $9\,m/s$. Find the motorcycle’s position, distance from the edge of the cliff and velocity after $0.5\,s$.

Answer 1

Hint: We can use the first and second equation of motion to find the answer to this question. Using the second equation of motion we can find the horizontal displacement and the vertical displacement at the given time. Thus, we will get the position. If we considered the end of the cliff as the origin. Distance from the origin to this point will give us the distance of the motor from the Cliff. Then using the first equation of motion we can find the vertical component of velocity and the horizontal component of velocity. Using this we can calculate the velocity of the motor at the time.

Complete step by step answer:
It is given that a motorcycle stunt rider rides off the edge of a cliff.
 The horizontal velocity at the edge is given as
${v_{ox}} = 9\,m/s$
We need to find the position, distance from the edge of the Cliff and the velocity after time
 $t = 0.5\,s$
Let us use the second equation of motion to find the position after $0.5\,s$
The second equation of motion is given as
$s = ut + \dfrac{1}{2}a{t^2}$
Where, s is the displacement, u is initial velocity, a is acceleration, t is the time taken.
First let us calculate the horizontal displacement.
The value of acceleration due to gravity in the horizontal direction is zero. Thus, we get
$x = {v_{0x}}t + \dfrac{1}{2}g{t^2}$
Initial velocity in x direction, ${v_{0x}} = 9\,m/s$
$ \Rightarrow x = 9 \times 0.5 + 0$
Now let us calculate the vertical displacement
.$y = {v_{0y}}t + \dfrac{1}{2}g{t^2}$
The initial velocity in y direction is zero because the velocity at the edge of the Cliff is horizontal so there is no velocity in the y direction in the beginning.
Hence, we get
$y = 0 + \dfrac{1}{2} \times 10 \times {0.5^2}$
$ \Rightarrow y = 1.25\,m$
If we take the end of the cliff as origin, $\left( {0,0} \right)$
 We can denote the position as $\left( {4.5, - 1.25} \right)$
The y coordinate is negative because at this time the motorcycle will be at a level lower than the starting point or the origin. That is, the point is lying in the negative y axis.
Now, distance of the point from the origin can be calculated using the distance formula.
The distance between the origin and the point can be found out as
$d = \sqrt {{{\left( {x - 0} \right)}^2} + {{\left( {y - 0} \right)}^2}} $
$ \Rightarrow d = \sqrt {{{\left( {4.5} \right)}^2} + {{\left( {1.25} \right)}^2}} $
$ \Rightarrow d = 4.67\,m$
Now let us find the velocity. For this, we can use the first equation of motion.
$v = u + at$
Where, v is the final velocity, u is the initial velocity, a is the acceleration and tis the time taken.
We the x component of velocity after $0.5\,s$ is
${v_x} = {v_{0x}} + gt$
$ \Rightarrow {v_x} = {v_{0x}} + 0$
Since g is equal to zero
$ \Rightarrow {v_x} = 9\,m/s$
The y component of velocity at the time $0.5\,s$ is
${v_y} = {v_{0y}} + gt$
$ \Rightarrow {v_y} = 0 + 10 \times 0.5$
$ \Rightarrow {v_y} = 0 + 10 \times 0.5$
$ \Rightarrow {v_y} = 5\,m/s$
Hence the resultant velocity at time $0.5\,s$ can be written as
$v = \sqrt {{v_x}^2 + {v_y}^2} $
On substituting the values, we get,
$ \Rightarrow v = \sqrt {{9^2} + {5^2}} $
$ \Rightarrow v = \sqrt {106} \,m/s$
This is the value of velocity after $0.5\,s$ .

Note: Remember that the acceleration due to gravity will always pull a body downward; it will never pull a body sideways. Hence whenever we consider the horizontal displacement the acceleration due to gravity will be zero. But there will be acceleration in the vertical direction.