Posted by matin on Wednesday, April 25, 2012 at 12:07pm.
A projectile is shot from the edge of a cliff 125 m above ground level with an initial speed of 105 m/s at an angle of 37.0o with the horizontal, as shown in Figure 339. (a) Determine the time taken by the projectile to hit point P at ground level. (b) Determine the range X of the projectile as measured from the base of the cliff. At the instant just before the projectile hits point P, find (c) the horizontal and vertical components of the velocity, (d) the magnitude of the velocity, and (e) the angle made by the velocity vector with the horizontal.

physics  Henry, Thursday, April 26, 2012 at 10:09pm
Vo = 105 m/s @ 37 Deg.
Xo = hor. = 105*cos37 = 83.9 m/s.
Yo = ver. = 105*sin37 = 63.2 m/s.
h = ho + (Y^2Yo^2)/2g.
h = 125 + (0(63.2)^2) / 19.6=328.8 m
above gnd.
Tr = (YYo)/g=(063.2) / 9.8=6.45 s.
= Rise time.
a. h = Vo*t + 4.9t^2 = 328.8 m.
0 + 4.9t^2 = 328.8.
t^2 = 328.8 / 4.9 = 67.1.
Tf = 8.19 s. = Fall time.
b. Range=Xo*(Tr+Tf)=83.9(6.45+8.19)=1228.3 m.
c. Y = Yo + gt.
Y = ver. = 0 + 9.8*8.19 = 80.3 m/s.
X = hor = Xo = 83.9 m/s.
d. V = sqrt(X^2+Y^2).
V = sqrt((83.9)^2+(80.3)^2) = 116 m/s.
e. tanA = Y/X = 80.3 / 83.9 = 0.95709.
A = 43.7 Deg.

physics  Lebron, Wednesday, October 7, 2015 at 12:09pm
How do you do part C

physics  curryhut, Sunday, June 12, 2016 at 3:05am
Part C
Vo = 105m/s theta= 37.0 degrees
Horizontal = 105*cos37 = 83.86m/s
Vertical = 105*sin37 = 63.19m/s
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