4.) In the metal ball drop demonstration done in class, a horizontally shot ball
ID: 1770907 • Letter: 4
Question
4.) In the metal ball drop demonstration done in class, a horizontally shot ball reached the floor in the same amount of time as one simply dropped from the same height. Explain why 5.) A golf ball is hit horizontaly from an elevated tee (vertically 256 feet higher than the hole). If the horizontal distance to the hole is 440 feet, find: a) the horizontal speed with which the bal was hit (110 s) b) the speed the ball will enter the hole with in both the horizontal () and verical () directions (110 fs 128 s) c) the speed and angle x·the bal wil enter the hole wth (see dagam) (168.8 fis; 49.2 6.) Yet another golf bal is sent flying at an angle of 30": the total time the ball spends in the air is 6 seconds. Use 10 m/s2 for the acceleration due to gravity. Find: a) the time for the ball to reach the top of its motion (3 s) b) the., of the ball (30 mis) c) the tofthe ball (60 mis) d)the % of the ball (S2 ms) e) the maximum height attained (45 m) fthe horizontal distance traveled (312 m ) Is there another angle that would have given you the same d, calcuiated in part f? h) What angle wold have given you the largest possible d, for the of the bal?Explanation / Answer
4) ball is shot horizontally so its initial vertical velocity, v0y = 0
for dropped ball, v0y = 0
for both ay = - g
yf- yi = v0y t + ay t^2 /2
everything is same so time taken to fall some height will be same for both.
5) in vertical,
v0y = 0
yf - yi = 0 - 256 = - 256 ft
- 256 = 0 - 32.174 t^2 /2
t = 3.99 sec
in this time, it will have to cover 440 ft in horizontal.
so v0 = 440 / 3.99
v0 = 110 ft/s .......Ans
(b) vx = v0x = 110 ft
vy = v0y - g t = 0 - (32.174)(3.99) = -128 ft/s
so vy = 128 ft/s donwwards
(c) Speed = sqrt(vx^2 + vy^2) = 169 ft/s
theta = tan^-1(vy/vx) = 49.3 deg below horizontal
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