Homework Clinic
Science Clinic => Physics => Topic started by: aero on Jun 23, 2013
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A projectile is launched from ground level to the top of a cliff which is 195 m away and 135 m high. If the projectile lands on top of the cliff 6.8s after it is fired, find the initial velocity of the projectile ( (a)magnitude and (b)direction ). Neglect air resistance.
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2D Kinematics so split into X and Y components:
X direction:
x = 195m
Vix = Vicos(theta)
Ax = 0 m/s^2
t = 6.8s
Y direction:
y = 135m
Viy = Visin(theta)
Ay = -9.8 m/s^2
Then we make two equations... one in x direction and one in y direction
You can isolate for a single variable.
(1) x = Vi*cos(theta)*t
x/tcostheta = Vi
(2) y = Vi*sin(theta)*t + 0.5 a*t^2
substitute (1) into (2) and we get:
(3) y = [x*sin(theta)*t]/[t*cos(theta)] + 0.5 a*t^2
you see that the time cancels and sin/cos is tan so we can simplify to
(3) y = x*tan(theta) + (1/2)a*t^2
plug in your numbers and solve for theta
theta = 61.66 degrees which is your direction then we could plug that back into (1) to solve for Vo.
(1) Vo = x/t*cos(theta) = (195) / (6.8)cos(61.66)
Vo = 60.4 m/s
Hope it's right!
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Height of Cliff = 135 = Vo Sin theta (t) - 1/2 g t^2 = Vo Sin theta (6,8) - 4.9 (6.8)^2
135 = 6.8 Vo Sin theta - 227
362 =6.8 Vo Sin theta
362/6.8 = Vo Sin theta = 53.2
Horizontal motion Vo Cos theta = 195
Tan theta = 53.2/195 and theta = 17 degrees
VoCos theta (6,8) = 195
Vo Cos theta = 28.7 m/s
Vo = 29.75 n/s
Vo Sin theta = 7.85 m/s