A flying man jumps off a cliff with a horizontal velocity of 30ms-1 and vertical upwayrd velocity of 10ms-1
Will he make the jump to the other side if the distance is 55metres? ( the cliff is the same height)
Note: Please answer in simple terms only. I am only just beginning physics so we have not yet done anything with angles etc. These are the equations we've learnt:
vf = at + vo
d= 1/2at^2
Will he make the jump to the other side if the distance is 55metres? ( the cliff is the same height)
Note: Please answer in simple terms only. I am only just beginning physics so we have not yet done anything with angles etc. These are the equations we've learnt:
vf = at + vo
d= 1/2at^2
-
There is a small change in the second equation it is not just 1/2at^2 but it is vo*t + 1/2at^2
First calculate the time of flight considering only the vertical velocity. Calculate the time taken to reach the maximum height
So vo = 10 ms-1 ( REMEMBER VERTICAL VELOCITY)
vf = 0 ms-1 (As the vertical velocity at the maximum height will be 0)
a = -9.81 (Acceleration due to gravity but sometimes it is taken as -10 or -9.8 for simplicity)
Now use the first equation to get the time taken and you get t = 1.019...s
Now calculate the max height reached using the second equation and you get d = 5.0968.....m
Assume that the man could reach the cliff. So he must have come down to a height of 5.0968.....m as the two cliffs are of the same height (The distance 5.0968...m is the vertical distance between the max height reached by the man to the cliffs)
So now you get
d = 5.0968
vo = 0 (as the man is starting to accelerate from a given height starting from rest. REMEMBER THAT WE ARE CONSIDERING ONLY VERTICAL VELOCITY)
vf = ? (here vf is the vertical speed at which the man reaches the next cliff)
a = 9.81
t = ?
Now calculate t using second equation and you get t = 1.019.....s
Now you have got the total time of flight.
Now come to the horizontal velocity
d = greater than 55m (It is obviously required to have a successful jump)
vo = 30 ms-1
vf = 30 ms-1 (As there is no force acting on the man to reduce his horizontal velocity)
a = 0
t = 1.019..... + 1.019...... = 2.038......s
Now if the man is successful in his jump then the above 5 details must be true
Now use the second equation and you get
d = 61.14....m
Therefore you can surely say that the jump would be successful.
OK IT'S DONE. HOPE YOU GOT IT
First calculate the time of flight considering only the vertical velocity. Calculate the time taken to reach the maximum height
So vo = 10 ms-1 ( REMEMBER VERTICAL VELOCITY)
vf = 0 ms-1 (As the vertical velocity at the maximum height will be 0)
a = -9.81 (Acceleration due to gravity but sometimes it is taken as -10 or -9.8 for simplicity)
Now use the first equation to get the time taken and you get t = 1.019...s
Now calculate the max height reached using the second equation and you get d = 5.0968.....m
Assume that the man could reach the cliff. So he must have come down to a height of 5.0968.....m as the two cliffs are of the same height (The distance 5.0968...m is the vertical distance between the max height reached by the man to the cliffs)
So now you get
d = 5.0968
vo = 0 (as the man is starting to accelerate from a given height starting from rest. REMEMBER THAT WE ARE CONSIDERING ONLY VERTICAL VELOCITY)
vf = ? (here vf is the vertical speed at which the man reaches the next cliff)
a = 9.81
t = ?
Now calculate t using second equation and you get t = 1.019.....s
Now you have got the total time of flight.
Now come to the horizontal velocity
d = greater than 55m (It is obviously required to have a successful jump)
vo = 30 ms-1
vf = 30 ms-1 (As there is no force acting on the man to reduce his horizontal velocity)
a = 0
t = 1.019..... + 1.019...... = 2.038......s
Now if the man is successful in his jump then the above 5 details must be true
Now use the second equation and you get
d = 61.14....m
Therefore you can surely say that the jump would be successful.
OK IT'S DONE. HOPE YOU GOT IT
-
The answer is YES.
I have uploaded the detailed solution at the following link as the equations cannot be written and diagram cannot be made here.
https://sites.google.com/site/physicsato…
Solution33.pdf
I have uploaded the detailed solution at the following link as the equations cannot be written and diagram cannot be made here.
https://sites.google.com/site/physicsato…
Solution33.pdf
-
Ah yeah he's flying