time in air:
hf=hi+vo*sinTheta*t-1/2 g t^2
0=0+3.5*sin30*t-4.8t^2
t(4.8t-3.5*1/2)=0
solve for t.
In an outdoor game a small bag of
rice is thrown from near the
ground with a velocity of 3.50 m/s
at an angle of 30.00 above the
horizontal. Find out how long the
bag will be in the air before it
lands at the same height from which it was thrown and how far it travelled horizontally.
2 answers
time till velocity (v=v0-at) is zero:
3.5/2 - 9.8t = 0
t = 1.75/9.8
That's how long it rose upward. It fell for the same time, so ...
3.5/2 - 9.8t = 0
t = 1.75/9.8
That's how long it rose upward. It fell for the same time, so ...
1 answer