projectile motion .help me with it
Attachments:
Answers
Answered by
1
Let the velocity with which it is projected is u
and the angle at which it is projected is A with the horizontal
So the y-component of the velocity vector is usinA
Total time of flight is 5+4=9sec
So in 9sec it falls on the ground again so displacement is 0 along y-axis
So considering only the y axis
0=usinA×9 -(9.8×9×9)/2
=>usinA=44.1 m/s
Now in 4 sec it has reached point P. Let P is at the height h from the ground so along y axis the displacement is h
h=usinA×4 - (9.8×4×4)/2
=(44.1)×4 - 78.4
=176.4+78.4
=254.8
and the angle at which it is projected is A with the horizontal
So the y-component of the velocity vector is usinA
Total time of flight is 5+4=9sec
So in 9sec it falls on the ground again so displacement is 0 along y-axis
So considering only the y axis
0=usinA×9 -(9.8×9×9)/2
=>usinA=44.1 m/s
Now in 4 sec it has reached point P. Let P is at the height h from the ground so along y axis the displacement is h
h=usinA×4 - (9.8×4×4)/2
=(44.1)×4 - 78.4
=176.4+78.4
=254.8
Endou:
you missed a point
its the case of projection from a height
but is it mentioned??
"it reaches the horizontal plane passing the point of projection "..
Similar questions