PLEASE HELP IM STRUGGLING HARDCORE!!! FIRST PERSON TO ANSWER GETS 10 POINTS ;P
A 0.4 kg rock is projected from the edge of
the top of a building with an initial velocity of
8.23 m/s at an angle 39 above the horizontal.
Due to gravity, the rock strikes the ground at
a horizontal distance of 14.7 m from the base
of the building.
How tall is the building?
A 0.4 kg rock is projected from the edge of
the top of a building with an initial velocity of
8.23 m/s at an angle 39 above the horizontal.
Due to gravity, the rock strikes the ground at
a horizontal distance of 14.7 m from the base
of the building.
How tall is the building?
-
We have to adopt a conversion technique here...s=-ut (plus) 0.5at^2....or here h= -usin@t (plus) 0.5gt^2...
We dont have the time..but we have vcos@*t=range or t=14.7/8.23 cos 39=2.298 s....
Plug up this value above to have h= -(8.23*sin 39*2.298) (plus) 4.9*2.298^2 ==> 13.97 m ..the plus sign indicates that the object is below the thrown height; as in above we took g as negative as it was also below; or acting downwards...hope this helps! ..cheers friend! :-)
We dont have the time..but we have vcos@*t=range or t=14.7/8.23 cos 39=2.298 s....
Plug up this value above to have h= -(8.23*sin 39*2.298) (plus) 4.9*2.298^2 ==> 13.97 m ..the plus sign indicates that the object is below the thrown height; as in above we took g as negative as it was also below; or acting downwards...hope this helps! ..cheers friend! :-)