Question

A staircase contains three steps each 10cm high and 20cm wide. What should be the minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly the lowest plane?

Answer 1

Here in the given situation we have to find the minimum velocity required for the ball to land on the lowest plane directly from the uppermost plane or stair. The ball will have a trajectory while having the motion. Considering the formula for the trajectory and redrawing the diagram we can find the minimum horizontal velocity of the ball.
Formula used:
$y=x\tan \theta -\dfrac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }$

Complete step-by-step solution:
From the data given in the question, we know the width and height of each stair. Now for minimum velocity, the ball just has to cover the distance of two stairs. As when it reaches the end of the second stair it will ultimately land on the last one. Hence the vertical displacement should be 20 cm and horizontal displacement should be 40cm.
Let us redraw the diagram for better understanding,

The trajectory for the vertical displacement is given as
$y=x\tan \theta -\dfrac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }$
Where x is the horizontal displacement, g is the acceleration due to gravity, u is the initial velocity and θ is the angle of elevation with respect to horizontal displacement.
Now, according to the question, the vertical displacement y and horizontal displacement x is given. g is the constant value whereas the angle of elevation will be zero as there is downward motion instead of upward motion so there won’t be any elevation and the motion is also in the horizontal direction.
As discussed earlier, for minimum velocity the vertical displacement should be 20cm (height 10cm each) and the horizontal displacement should be 40cm (two stairs of width 20cm each). Value of g is $9m/{{s}^{2}}$and $\theta =0$. The value of x in meters will be 0.4m and the value of y in meters will be 0.2m.
Substituting the value in the above equation we get

& -\left( 0.2 \right)=\left( 0.4 \right)\tan 0-\dfrac{(9.8){{\left( 0.4 \right)}^{2}}}{2{{u}^{2}}{{\cos }^{2}}0} \\\ & \Rightarrow -0.2=0-\dfrac{(9.8){{\left( 0.4 \right)}^{2}}}{2{{u}^{2}}{{(1)}^{2}}}\text{ }\left( \because \tan 0=0\text{ and}\cos 0=1 \right) \\\ & \Rightarrow -0.2=-\dfrac{(9.8){{\left( 0.4 \right)}^{2}}}{2{{u}^{2}}} \\\ & \Rightarrow {{u}^{2}}=\dfrac{\left( 9.8 \right){{\left( 0.4 \right)}^{2}}}{2\times 0.2} \\\ & \Rightarrow {{u}^{2}}=3.92 \\\ & \Rightarrow u=\sqrt{3.92} \\\ & \Rightarrow u=1.97 \\\ & \Rightarrow u\simeq 2m/s \\\ \end{aligned}$$ The value of y is taken negative as the motion is downward. **Hence the minimum horizontal velocity of ball should be approximately $$2m/s$$.** **Note:** In case the ball was thrown in upward direction there will be some value of angle of elevation. Also, note that y is vertical displacement and not distance therefore it has direction also and we have to consider the negative sign to show the direction. Changing units in the SI unit is necessary because if there is a slight change in decimal place the answer would change as it is the square root.