A stone is thrown vertically upwards with a velocity of 19.6 m/s.After 2 sec,another stone is thrown upwards with a velocity of 9.8 m/s.When and where will these stones collide?
Answers
Answer:
firstly we will calculate the distance travelled by the first stone which thrown with velocity 19.6 in 2 second (In 2 second this stone will come to a state of momentarily rest) and it comes out to be 19.6 meter . after the ball comes to a state of momentarily rest then second ball is thrown which comes to rest after 1 second and has travelled a distance of 4.9 meter in that 1 second first ball has also travelled a distance of 4.9 meter and has gained a velocity of 9.8 m\s (s=ut+0.5gt^2 where u is 0 as it is the velocity when ball is at momentarily rest ,t is time =1 second)the distance between two balls comes out to be 9.8 and the two balls has travelled for 3 second considering that at this point observer has started measuring the time we can calculate the fraction of time required for collision assuming that ball collide at X distance above ground and at time T we can calculate the velocity of stones at that point by using two equations for each stone that is V^2=U^2+2gs ( for first stone U^2=9.8^2 ,s=14.7-X and for second stone U^2=0,s=4.8-X) and V=U+gt(for first U=9.8 ,t=t and for second U=0,t=t)
solving the above we will end up with 4 equation 2 for stone first and 2 for stone second putting the 2 equations of 1st stone as equal to each other as both will give velocity of first stone at the point where the two stone will collide and 2 equations of second stone as equal to each other as both will give velocity of second stone at the point where two stone will collide after putting the equations equal we are left with 2 equations and by solving it simultaneously we can evaluate the value of t whisch comes out to be 1 thiu total time taken is T = 2+1+1=4
You can also use the concept of relative velocity which will make the solution short and you will avoid some of the calculations.
Thank you❤❤❤