Trajectory in Uniform G-field Points: 1 The trajectory of a rock thrown from a h

Question

Trajectory in Uniform G-field Points: 1 The trajectory of a rock thrown from a height with an initial speed of 20.5 m/s is shown in the figure below. Evaluate the magnitude of the gravitational field at 50 F 45 40 35 30 25 15 10 0 10 20 040 50 60 70 80 90 100 110 Horizontal Position (n) For a given initial velocity the initial angle is related to the "range" and "rise" of a projectile To evaluate the initial slope: It may be easier to find a tangent on the way down instead of on the way up Submit Answer Incorrect. Tries 4/10Previous Tries

Explanation / Answer

x direction

110 = 20.5*cos(theta) t

t = 110/(20.5*cos(theta))

y direction

max height = 50 m

v^2 = v0^2 + 2 a y

0 = (20.5*sin(theta))^2 - 2*a*(50-15)

a = (20.5*sin(theta))^2/ 70

y = y0 + v0y t + 1/2 a t^2

0 = 15 + (20.5*sin(theta))*t - 0.5*a*t^2

0 = 15 + (20.5*sin(theta))*(110/(20.5*cos(theta))) - 0.5*((20.5*sin(theta))^2/70)*(110/(20.5*cos(theta)))^2

solve for theta

theta = 0.95 radians

so

a = (20.5*sin(0.95))^2/70 = 3.97 m/s^2

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