PHYS 193 Spring 2016 Final Exam Write your answers (with appropriate precision)

Question

PHYS 193 Spring 2016 Final Exam Write your answers (with appropriate precision) in the blanks provided. Indicate all units. 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 from that question here). Name: x=mt + b ; m=4.9 m/s; b=-0.026 rn Fill out the table below , y=At+ Bt+C; A=-5.2 ; B=3.5 ; C=-0.049 Initial v Initial 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 Vx at the top Vy at the top Range of the projectile Sketch thex and y- position graphs for this motion. Make sure that the graph shows a) whether V or Vy is larger b) the time the max height was reached, and what that max height was c) the time the range (maxx) was reached, and what that range was 1.6 1.4 1.2 0.8 0.6 0.4 0.2 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 4

Explanation / Answer

4. given

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 the following are the values

Initial Vx = x'(t = 0) = m = 4.9 m/s

Initial Vy = y'(t = 0) = B = 3.5 m/s

Initial Magnitude of V = sqroot(Vx^2 + Vy^2) = 6.0216 m/s

Launch angle = theta = arctan(Vy/Vx) = 35.537 deg

g = y" = 2A = -10.4 m/s/s

%error = (10.4 - 9.81)*100/9.81 = 6.01427 %

Time to the top = t

vy = 0 at top, hence, 2At + B = 0

2(-5.2)*t + 3.5 = 0

t = 0.3365 s

Maximum height = y(t) = 0.5398 m

Vx at top = m = 4.9 m/s

Vy at top = 0

Range , x(2t) - x(0) = 3.2977 m

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