Answer to Question #97819 in Mechanics | Relativity for C

Question

Stone A is thrown vertically upwards with a speed of 10 meters per second from the edge of the roof of a 40m high building.Ignore the effects of friction.Take the ground as reference.
First we had to calculate maximum height then,please help me with these questions?with detailed explanation of every step please?And equations and calculations please?
1)Write down the magnitude and direction of the acceleration of stone A at this maximum height.
2Now Stone B is dropped from rest from the edge of roof x seconds after stone A was thrown upwards.Stone A passes stone B when the two stones are 29,74m above the ground.Calculate the value of x.

Answer 1

Maximum height = $\frac{u^2}{2g}=\frac{100}{20}=5\ m$ from the edge of roof

1.)

At every instant the magnitude and direction of acceleration remain same

So, at maximum height aceleration will be $10\ m\ sec^{-2}$ towards downward

2.)Let downward direction to be positive

For stone A,

$$10.26=(-10)t+0.5\times10t^2=-10t+5t^2\ \ \ .....(1)$$

Solving it by srithacharya formula,

$$t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{10\pm\sqrt{100+20\times10.26}}{10}=2.747\ sec$$

For Stone B,

$$10.26=0.5\times10(2.747-x)^2\\or\\2.052=(2.747-x)^2\\or\\1.432=(2.747-x)\\or\\x=2.747-1.432=1.315\ sec$$