I just need the one I got wrong A shot putter releases the shot some distance ab
ID: 585255 • Letter: I
Question
I just need the one I got wrong A shot putter releases the shot some distance above the level ground with a velocity of 10.8 m/s, 49.0 degree above the horizontal. The shot hits the ground 2.20 s later. You can Ignore air resistance. (Assume the horizontal direction of the shot and upward are positive.) (a) What are the components of the shot's acceleration while in flight? (b) What are the components of the shot's velocity at the beginning of its trajectory? What are the components of the shot's velocity at the end of its trajectory? (c) How far did she throw the shot horizontally? (d) Why does the expression for R (as given below) not give the correct answer for part (c)? (e) How high was the shot above the ground when she released it? (f) Draw x - t, y - t, v_x - t and v_y - t graphs for the motion.Explanation / Answer
Finding the y component of the velocity which is (10.8sin49.4).
You then know:
a = -9.81ms^-2 (opposite direction thus negative)
u = (10.8sin49.4)ms^-1
t = 2.2 s
s = ?
You then put these value into the kinimatic equation:
s = ut + 0.5a(t^2)
s = 3.5 m above the ground level
H = Maximum Height of projectile
H = (10.8sin49.0)^2/2×9.8
H = 3.38m
Total maximum height = 3.5 + 3.38 = 6.88 m
Time for maximum height = (10.8sin49.4)/9.8
T = 0.83 s
Hence time for down motion = 2.2-0.83 = 1.37 s
Therefore velocity component at the end of the shot = 9.8×1.37 = 13.42 m/s^2