in the blanks provided. Indicate all units Write your answers (with appropriate

Question

in the blanks provided. Indicate all units Write your answers (with appropriate precision) 4. In a "Projectiles" experiment the two curves were fitted to the equations (Please note: this question is entirely separate from question 1. These data were obtained for a projectile launched at a different angle with a different initial velocity It would make no sense to use any results use any results from that question here) x=mt + b ; m=4.9m/s ; b=-0.026 m Fill out the table below y=At2 + Bt+C; A=-52; B=3.5;C=-0.049 nitial v nitial v Magnitude of initial v Launch angle % error in the value of g (theoretical value g 9.8 m/s) Time to the top Maximum height V, at the top Vy at the top Range of the projectile Sketch the x and y- position graphs for this motion. Make sure that the graph shows a) whether v, or v, is larger bj the time the max height was reached, and what that max height was c) the time the range (max x) was reached, and what that range was 1 6 1.4 12 0 8 0.6 0.4 0 2 0 0 1 0.2 03 0 4 05 0.6 0.7 0.8

Explanation / Answer

for the given problem

x = mt + b
m = 4.9 m/s
b = -0.026 m

y = At^2 + Bt + C
A = -5.2
B = 3.5
C = -0.049

hence
vx = dx/dt = m
vx initial = 4.9 m/s

vy = dy/dt
intiail vy = B = 3.5 m/s

initial v = sqroot(vx^2 + vy^2) = 6.0216 m/s

Launch angle = arctan(vy/vx) = 35.54 deg

g = 2A = -10.4 m/s/s
%error = (10.4 - 9.81)*100/9.81 = 6.014 %

time to top = t
vy = 2At + B = 0
t = -B/2A = 0.33653846 s

maximum height = y(t) = 0.5399423 m

Vx at top = m = 4.9 m/s
Vy at top = 0 m/s
range = x(2t) - x(0) = 3.298 m

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