In Denver, children bring their old jack-o-lanterns to the top of a tower and co
ID: 1377944 • Letter: I
Question
In Denver, children bring their old jack-o-lanterns to the top of a tower and compete for accuracy in hitting a target on the ground. Suppose that the tower height is h = 7.80m, and that the bulls-eye's horizontal distance is d = 3.3 m from the launch point. (Neglect air resistance.)
If a jack-o-lantern is given an initial horizontal speed of 3.3 m/s, what are the direction and magnitude of its velocity at the following moments in time? (Let the +x axis point to the right.)
(a) 0.75 s after launch
Your response differs from the correct answer by more than 10%. Double check your calculations.
Explanation / Answer
Vxi = 3.3m/s...it never changes
Vyi = gt, where g = gravity = 9.8m/s^2 and t = time.
V = SQRT(Vxi^2 + Vyi^2)
V is the actually magnitude of the velocity.
To find the angle:
? = arcsin(Vyi/V)
? = arccos(Vxi/V)
? = arctan(Vyi/Vxi)
You can use any of these three formulas. They give the same answer.
(a)
Vyi = 9.8(.75) = 7.35m/s
Vxi = 3.3m/s
V = SQRT(3.3^2 + 7.35^2) = 8.05m/s
? = arctan(7.35/3.3) = 65.93