Question

From the top of a building of height h = 100 m I throw a stone up with velocity 10 m/s. What is
the maximum height it reaches, and when does this occur? How many seconds does it spend on
its way down between h = 50 m and h = 0 m? What is its velocity when h = 50 m? If, while the
stone is airborne, an earthquake opens up a hole 50 m deep in the ground, when and with what
speed will the stone hit the bottom?

There are answers but I would like if someone would like to work together and explain each process.

Answer 1

If you take g = 10 m/s^2 for this question, it may be helpful to understand the logic

Answer 2

Just think that when you throw up the stone, when will it stop ? You throw it with 10 m/s upwards (+y) and there is gravitational acceleration (g) downwards (-y) so -10m/s^2, then it takes just a second for the stone to stop. For the first second it should take 5 meters (h=1/2*g*t^2), and that's all since it is just a second. Now that it reached h = 105 m, after this point applies free fall. You accelerate 10 m/s then it should take around (105-50=1/2*10*t^2 ==> t= 3.32) seconds to reach h=50 m, and 4.6 seconds to reach h = 0 m. Since the final velocity is just the initial velocity + g*t, v should be 0 + 10*(3.32) for h = 50 m. When a hole is opened deep in the ground then this basically means that h is now 100+50 = 150 m (before throwing).

Answer 3

first use V=U+AT here take downward direction +ve (u may take the opposite it's totally upto you)then gravitational acceleration will be +ve and the velocity will be negative.hence time for upward flight will be 0=-10+10t i.e. t=1sec.we can use s=ut+0.5at^2 we will get s as -5 (negative sign due to sign convention) hence max height attained by the particle will be 100+5=105m.now after traveling downwards for 1 sec the velocity of the particle will be 10m/s.now again follow the same procedure use 2nd kinem.eqn u will get 100=10t+5t^2

Answer 4

;)