Question

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 3-39. (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.

Answer 1

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^2-Yo^2)/2g.
h = 125 + (0-(63.2)^2) / -19.6=328.8 m
above gnd.

Tr = (Y-Yo)/g=(0-63.2) / -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.

Answer 2

How do you do part C

Answer 3

Part C
Vo = 105m/s theta= 37.0 degrees
Horizontal = 105*cos37 = 83.86m/s
Vertical = 105*sin37 = 63.19m/s