A stone is thrown from the top of a building with an initial...

A stone is thrown from the top of a building with an initial velocity of 20.0 m/s straight upward. The building is 50.0 m high, and the stone just misses the edge of the roof on its way down. a) Determine the time needed for the stone to reach its maximum height b) Determine the maximum height c) Determine the time needed for the stone to return to the level of the thrower d) Determine the velocity of the stone at this instant e) Determine the velocity and position of the stone at t= 5.00 s f) Determine the velocity of the stone just before it hits the ground

Transcript

00:01 I have a question and this it is given that a stone is thrown from top of a building with an initial velocity of 20 meter per second if this is top of a building it was being thrown with initial velocity of 20 meter per second upward the building is 50 meters high building is 50 meters high and the stone just misses the edge of the roof and its way down which amount the time needed for the stone to reach its maximum height and determine the maximum height okay the velocity of the stone okay various things we need to find out first is i have to find out the time we did for a stone to reach maximum height we know that at maximum height we should be equal to zero this is h and g will always be acting vertically downward so be equal to v square equal to u sq plus 2 s so 0 will be equal to u squared minus 2g h will be equal to u s squared by 2g u s square is 20 square that is 400 by 2 into if we take g as 10 this would be 20 maximum height will be 20 meters okay remember we have taken g equal to 10 meter per second square if i would have taken it 9 .8 then it could have been more than this part b part b okay this is a maximum height okay and maximum height is this is for part b answer should be a part b maximum height from the roof this is a part b sorry for this and find this i'll find that now the time taken so time taken we can take be equal to u plus 80 so time will be equal to minus u by minus g so 20 by 10 that is two seconds okay this is a two seconds and remind the maximum height only taken time needed for stone to return to the level of form remind the velocity the at this instant velocity of the stone when it was returning okay that is velocity of the stone after four seconds or after two seconds if we assume from top position from top position initial velocity you should be equal to zero height is 20 meters and acceleration is 10 meter per second square for sake of simplicity the velocity will be equal to b squared equal to u square plus 2 into b into h so v will be equal to 20 meter per second will be the speed when returning and determine the velocity of the stone at this instant and determine the velocity at the position when d equal to five second determine the velocity of the stone just before it hits the ground in velocity of storm after five seconds we'll be using the formula v equal to u plus 80 after four seconds it has velocity 20 meter per second so after five seconds it will have velocity 20 plus 10 that is 30 meter per second as velocity is increasing with 10 meter per second square he demand the velocity of the stone just before it hits the ground okay the total height will be equal to 70 meters while traveling at the top u equal to zero h is this and acceleration due to gravity is 10 meter per second square velocity b will be equal to 2 into 10 into 70 and the root this is the velocity with which it will hit the ground and the root 1 is 4 37 .42 meter per second...

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