NOTE: Take UP as POSITIVE. A rock is projected from the edge of the top of a bui

Question

NOTE: Take UP as POSITIVE.
A rock is projected from the edge of the top
of a building with an initial velocity of 24 m/s
at an angle of 31? above the horizontal. The
rock strikes the ground a horizontal distance
of 84 m from the base of the building.
The acceleration of gravity is 9.8 m/s2 .
Assume: The ground is level and that the
side of the building is vertical. Neglect air
friction.
a) what is the horizontal component of the
rock’s velocity when it strikes the ground?
Answer in units of m/s.

b) How long does the rock remain airborne? Answer in units ofs.

c)How tall is the building? Answer in units of m.

d)What is the vertical component of the rock’s
velocity when it strikes the ground? Answer
in units of m/s.

e)What is the magnitude of the rock’s velocity
when it strikes the ground? Answer in units
of m/s.

Explanation / Answer

   initial velocity, u = 24 m/s    angle of projection, = 31o    horizontal range, X = 84 m (a) horizontal component of the rock’s velocity when itstrikes the ground = u cos = 20.57 m/s (b) time of travel, t = X / ( u cos ) = 84 /20.57 = 4.08 s (c)    - h =  ( u sin ) t - (1/2 ) g t 2            h= (1/2) gt 2 -  ( u sin ) t =(1/2) (9.8) (4.08)2 - (12.36) (4.08) = 81.567 - 50.4288= 31.1382 m (d)   Vertical component, Vy = ( u sin )- g t = 12.36 - (9.8) (4.08) = -27.624 m/s ( downward) (c) Final velocity, V = [ (Vx) 2 + (Vy)2] 1/2 = [ ( u cos) 2 + (Vy)2 ] 1/2                                                                      =  [( 20.57) 2 + (27.624) 2 ]1/2 = 34.44 m/s                                                                      =  [( 20.57) 2 + (27.624) 2 ]1/2 = 34.44 m/s

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