00:01
Hello student, to solve the given question, let us write the equation of motion that is s equals ut plus half a t square.
00:09
Also we know the value for x, that is the horizontal distance, equals velocity v, multiply by time t.
00:16
So using these relations, let us solve for part a, that is, using the equation of motion, we get minus h equals 0 plus half minus g multiplied by t squared.
00:29
So from here we get the value for t equals 2h divided by g under root.
00:35
Mark it as equation number 1 for further use.
00:38
Also plug in the values, t will be equal to 2 multiply by 78 .4 divided by 9 .8 underrout.
00:47
So from here we get the final answer for part a, that is, t equals 4 seconds.
00:55
Now moving on to the part b, we need to find the displace.
00:59
For this case x will be equal to v is 5 multiplied by t which is 4 it will be equal to 20 meters similarly for part c we need to firstly find the horizontal component that is vx that is already 5 .0 meters per second and the vertical component v y is minus g that is 9 .8 multiply by time 4 it will be equal to minus 39 .2.
01:31
So the final answer for part b and c, x is equal to 20 meters, while for part c, vx is 5 .0 meters per second and vy is minus 39 .2 meters per second.
01:51
Now furthermore, according to the given conditions, for part 1, if the velocity is, that is, v -dash is double, that is the 2v, hence it will be equal to 2 into 5 equals 10 meters per second...