Asked by Physics
a man stand on a 100 m cliff and throws a rock with speed of 40ms at 42degrees above horizontal.
what time before rock hits ground.
the horizontal distance the rock traveled from base of cliff.?
my answer is 4.08 sec before rock hits ground
81.6 m distance traveled from base of cliff
what time before rock hits ground.
the horizontal distance the rock traveled from base of cliff.?
my answer is 4.08 sec before rock hits ground
81.6 m distance traveled from base of cliff
Answers
Answered by
drwls
The height of the rock above the base of the cliff is:
Y = 100 + 40 sin 42 t - 4.905 t^2
= 100 + 26.765 t - 4.905 t^2.
When t = 4.08, Y = 127 m
Your answer is incorrect.
Solve for t when Y = 0
Then multiply that time by 40 m/s*cos 42, the horizontal velocity component.
Y = 100 + 40 sin 42 t - 4.905 t^2
= 100 + 26.765 t - 4.905 t^2.
When t = 4.08, Y = 127 m
Your answer is incorrect.
Solve for t when Y = 0
Then multiply that time by 40 m/s*cos 42, the horizontal velocity component.
There are no AI answers yet. The ability to request AI answers is coming soon!