Posted by Leo on .
Hello everyone, I need help with this question:
A dart gun is fired while being held horizontally at a height of 1.00m above ground level and while it is at rest relative to the ground. The dart from the gun travels a horizontal distance of 5.00m. A college student holds the same gun in a horizontal position while sliding down a 45.0 degree incline at a constant speed of 2.00m/s. How far will the dart travel if the student fires the gun while it is 1.00m above the ground?
Thanks very much!
As far as it wants! Your most welcome! :-)
=sqrt(2•1/9.8) =0.45 s
s=v(x) •t =>
The dart initial speed is
v(x) = s/t =5/0.45=11.07 m/s.
(b) v(y) =v•sin45º=2•0.707 =1.41m/s
h=v(y) •t +gt²/2
gt²/2 + v(y) •t –h =0,
4.9 t² +1.41•t -1 =0
t1=0.33 s. t2 - negative root
s1=v(x) •t1= 11.07•0.33=3.66 m.
Elena's answer is wrong
last step should be 12.48(0.33)
answer is 4.12m
the time is supposed to be given , it was 0.329