A projectile is shot from the edge of a cliff 115 {\ m m} above ground level wit
ID: 2216148 • Letter: A
Question
A projectile is shot from the edge of a cliff 115 { m m} above ground level with an initial speed of 65.0 { m{ m/s}} at an angle of 35.0 { m{^circ }} with the horizontal, as shown in the figure . a) Determine the time taken by the projectile to hit point p at ground level b) Determine the distance x of point p from the base of the vertical cliff. At the instant just before the projectile hits point p,find c) the horizontal and the vertical component of its velocity d)the magnitude of the velocity and e) the angle made by the velocity vector with the horizontal f) find the maximum height above the cliff top reached by the projectileExplanation / Answer
a) Let t = time taken by the projectile to hit point P at ground level in
second
dy = Viy(t) + (1/2)(g)(t^2)
dy = (Vosin)(t) + (1/2)(g)(t^2)
- 115 m = (65 m/s x sin 35.0°)(t) + (1/2)(-9.80m/s^2)(t^2)
- 115 = 37.28t - 4.9t^2
4.9t^2 - 37.28t - 115 = 0
Using the quadratic formula we get...
t = 9.96 sec answer
b) X = range of projectile as measured from the base of the cliff in km.
X = (Vix)(t)
X = (Vocos37°)(t)
X = (65 m/s x cos 35°)(9.96 s)
X = 530.3 m or 0.53 km answer
c) horizontal component = Vx = Vix = 65 m/s x cos 35° = 53.24 m/s
Vx = 53.24 m/s answer
vertical component = Vfy = sqrt[(2gdy) + Viy^2]
Vfy = sqrt[2(-9.8)(-115) + (65 x sin35°)^2]
Vfy = 60.36 m/s answer
d) Vr = magnitude of the velocity
Vr = sqrt[Vx^2 + Vfy^2]
Vr = sqrt[(60.36)^2 + (53.24)^2]
Vr = 80.48 m/s answer
e) = angle made by the velocity vector below the horizontal
= arctan(Vfy/Vx)
= arctan(60.36/53.24)
= 48.5885° answer
f) dymax = maximum height above the cliff top reached by the
projectile in m.
2g(dymax) = Vfy^2 - Viy^2
dymax = (Vfy^2 - Viy^2)/(2g)
dymax = [0^2 - (65 x sin 35°)^2]/[2(-9.8)]
dymax = 70.9174 m answer
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