Asked by Joanne
a projectile is fired at 50 meters per second at an angle of 30 degrees below the horizontal off a cliff that is 50 meters high. How far away and at what speed and angle will this projectile land?
Answers
Answered by
Henry
Vo = 50m/s[30o]
Xo = 50*Cos(30) = 43.3 m/s.
Yo = 50*sin(30) = 25 m/s.
h = Yo*t + 0.5g*t = 59 m.
25t + 4.9t^2 = 50
4.9t^2 + 25t - 50 = 0
Use Quadratic Formula.
Tf = 1.54 s. = Fall time.
Dx = Xo * Tf = 43.3m/s * 1.54s = 66.7 m.
From the cliff.
Y = Yo + g*t = 25 + 9.8*1.54 = 40.1 m/s.
= Y component of Landing speed.
Tan A = Y/Xo = 40.1/43.3 = 0.92523
A = 42.8o = Landing angle.
V = Y/sin A = 40.1/sin42.8 = 59 m/s. =
The landing velocity.
Xo = 50*Cos(30) = 43.3 m/s.
Yo = 50*sin(30) = 25 m/s.
h = Yo*t + 0.5g*t = 59 m.
25t + 4.9t^2 = 50
4.9t^2 + 25t - 50 = 0
Use Quadratic Formula.
Tf = 1.54 s. = Fall time.
Dx = Xo * Tf = 43.3m/s * 1.54s = 66.7 m.
From the cliff.
Y = Yo + g*t = 25 + 9.8*1.54 = 40.1 m/s.
= Y component of Landing speed.
Tan A = Y/Xo = 40.1/43.3 = 0.92523
A = 42.8o = Landing angle.
V = Y/sin A = 40.1/sin42.8 = 59 m/s. =
The landing velocity.
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