Question

A small ball is released from rest and falls on the horizontal platform which is

descending with a constant speed 7 1ms-1 .Given that the ball is 12m above the

platform at the instant of release. Calculate the time that elapses before the ball

hits the plat form.

Answer 1

As per the given question,

Initial height from the floor of the platform $(h_1)=12m$

Speed of the platform $(v)=71 m/sec$

Let after time t, both will be at the same height,

Hence, distance covered by the platform in t time is x,

$x=71\times t$

Now, height covered by the ball in the same time t will be,

$12+x=ut+\dfrac{gt^2}{2}$

here the initial velocity of the ball is u and g is the gravitational acceleration,

$g= 9.8 m/sec^2$

now substituting the values in the above,

$12+71t=0+\dfrac{9.8\times t^2}{2}$

$\Rightarrow 12+71t=4.9t^2$

$\Rightarrow 4.9t^2-71t-12=0$

we know that, if $ax^2+bx+c=0$

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Hence

$t=\dfrac{71\pm\sqrt{71^2+4\times 4.9\times12}}{2\times 4.9}$

$t=\dfrac{71\pm 72.63}{9.8}$

We know that time, never can be negative, hence leaving the negative value.

$t=\dfrac{71+72.63}{9.8}=14.65 sec$