A projectile is shot from the edge of a cliff h = 125 m above ground level with

Question

A projectile is shot from the edge of a cliff h = 125 m above ground level with an initial speed of v0 = 125 m/s at an angle of 37.0° with the horizontal, as shown in Fig. 3-39.

(a) Determine the time taken by the projectile to hit point P at ground level.
1: ________ s
(b) Determine the range X of the projectile as measured from the base of the cliff.
2: __________ km
(c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity (take up and to the right as positive directions).
3:  ___________  m/s (horizontal component)
4: ____________ m/s (vertical component)
(d) What is the the magnitude of the velocity?
5: _____________ m/s
(e) What is the angle made by the velocity vector with the horizontal?
6: ______________° (below the horizontal)

Explanation / Answer

vo=125 m/s==>vox=vo*cos(37º)99.83 m/s==>voy=vo*sin(37º)75.227 m/s

yf=yi+voy*t+1/2*ay*t²==>vyf(voy²+2*g*125)90.1 m/s==>vf(90.1²+99.83²)134.44 m/s

max ht=voy²/(2*g)75.227²/(2*9.8)(288.73 +125)m413.73 m

max ht t=voy/g=75.227/9.87.68 s times 2 to return to level with cliff15.3524 s

t from cliff to grd level=(voy+(voy²+2*g*y))/g(75.227+(75.227²+2*9.8*125))/9.816.865 s

total time=(16.865+15.3524)s32.22 s

a) 32.22 s

b) 32.22*99.833216.3 m

c) vox-99.83 m/s==>voy-90.1 m/s

d) vf134.44 m/s

e) arctan(-90.1/99.83)=-42.1º

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