Suppose a stone is thrown from 45.0 m above the ground at anangle of 30 degrees

Question

Suppose a stone is thrown from 45.0 m above the ground at anangle of 30 degrees below the horizontal. If it strikes the ground57.0 m away, find a) the time of flight b) the intial speed, and c)the speed and the angle of the velocity vector with respect to thehorizontal at impact. This is Exercise 3.8 on page 69 in the College Physics 8THediton. Suppose a stone is thrown from 45.0 m above the ground at anangle of 30 degrees below the horizontal. If it strikes the ground57.0 m away, find a) the time of flight b) the intial speed, and c)the speed and the angle of the velocity vector with respect to thehorizontal at impact. This is Exercise 3.8 on page 69 in the College Physics 8THediton.

Explanation / Answer

height, h = 45 m angle of projection, = 30o horizontal range, X = 57 m initial velocity of projection = U horizontal component of initial velocity, Ux = U cos Vertical component of initial velocity, Uy = U sin horizontal range, X = ( U cos ) t 57 = ( U cos ) t t = 57 / ( U cos )           ............................(1) (b) S = ut + ( 1/2) at2 - h = - ( U sin ) t - ( 1/2) gt2 h = ( U sin ) t + ( 1/2 ) gt2 From equ.(1) h = ( U sin ) [ 57 / ( U cos ) ] + ( 1/2)g [ 57 / ( U cos ) ]2 45 = 57 tan 30o + 4.9 * 4332 / U2 21226.8 / U2 = 12.09 U = 41.9 m/s (a) (a) time of flight, t = 47 / ( U cos ) = 57 / ( 41.9 * cos30o ) = 1.57s (c) horizontal component of final velocity, Vx = Ux = U cos = 36.29 m/s Vertical component of final velocity, Vy = U sin + gt= -36.34 m/s Final velocity, V = [ ( Vx ) 2 + ( Vy )2 ] 1/2 = 51.36 m/s angle at which reach the ground, = arc tan ( Vy / Vx )= -45o ( below horizontal ) (c) horizontal component of final velocity, Vx = Ux = U cos = 36.29 m/s Vertical component of final velocity, Vy = U sin + gt= -36.34 m/s Final velocity, V = [ ( Vx ) 2 + ( Vy )2 ] 1/2 = 51.36 m/s angle at which reach the ground, = arc tan ( Vy / Vx )= -45o ( below horizontal )

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