A spring cannon is located at the edge of a table that is 1.20 m above the floor

Question

A spring cannon is located at the edge of a table that
is 1.20 m above the floor. A steel ball is launched from the
cannon with speed vi at 35.0° above the horizontal. (a) Find
the horizontal position of the ball as a function of vi at the
instant it lands on the floor. We write this function as x(vi ).
Evaluate x for (b) vi = 0.100m/s and for (c) vi =100 m/s.
(d) Assume vi is close to but not equal to zero. Show that
one term in the answer to part (a) dominates so that the
function x(vi) reduces to a simpler form. (e) If vi is very
large, what is the approximate form of x(vi)? (f) Describe
the overall shape of the graph of the function x(vi ).

Explanation / Answer

?h= -1.2 = Vosin35*t - 1/2*g*t*t x = Vocos35 * t now, find t from first equation in terms of Vo , gusing quadratic formula. substitute in secod equation to find x. (g/2)*t2 - Vosin35*t - 1.2 = 0 => t = [Vosin35 ±v{Vo2sin235 + 4.8g}] / g since in second case { where - is taken } t< 0 , we take : t = [Vosin35 +v{Vo2sin235 + 4.8g}] / g so, x(Vo) = Vocos35 * [Vosin35 +v{Vo2sin235 + 4.8g}] / g when Vo ~ 0 , Vosin35 +v{Vo2sin235 + 4.8g} ~~ v(4.8g) so, x(Vo) ~ Vocos35v(4.8/g) if Vo is very large, Vosin35 +v{Vo2sin235 + 4.8g}] ~~~ 2Vosin35 so, x(Vo) ~~~ 2Vo2sin35cos35 / g = Vo2sin70/ g

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