Suppose the stone is thrown at an angleof 25.0° below the horizontal fromthe sam

Question

Suppose the stone is thrown at an angleof 25.0° below the horizontal fromthe same building (h = 50.0m) as in the example above. If it strikes the ground 91.8 m away, find the following.(Hint: For part(a), use the equation for the x-displacement to eliminatev0t from the equation for they-displacement.) (a) the time of flight 5 s

(b) the initial speed 6 m/s

(c) the speed and angle of the velocity vector with respect to thehorizontal at impact speed 7 m/s
angle 8° below the horizontal Suppose the stone is thrown at an angleof 25.0° below the horizontal fromthe same building (h = 50.0m) as in the example above. If it strikes the ground 91.8 m away, find the following.(Hint: For part(a), use the equation for the x-displacement to eliminatev0t from the equation for they-displacement.) (a) the time of flight 5 s

(b) the initial speed 6 m/s

(c) the speed and angle of the velocity vector with respect to thehorizontal at impact speed 7 m/s
angle 8° below the horizontal (a) the time of flight 5 s

(b) the initial speed 6 m/s

(c) the speed and angle of the velocity vector with respect to thehorizontal at impact speed 7 m/s
angle 8° below the horizontal speed 7 m/s
angle 8° below the horizontal

Explanation / Answer

Along the vertical direction :                y-yo = (vosin ) t -0.5gt2               -50.0m = (vosin25o ) t -  4.9t2                    -118.3m= vo t -  11.59t2       .........1 Along the horizontal direction :                  x - xo = (vocos ) t                  101.29 m =vot                  ........2 Solve eq 1 and eq 2 for time of flight 't'.               -118.3m = 101.29 m-  11.59t2                    t = 4.35s             ......3 Substituting 3 in eq 2 we get Initial velocity vo = 23.28 m/s The horzontal component of velocity vx does notchange from its initial value because thee is no horizontalacceleration. Thus, when the stone reaches the ground                          vx = vo cos = 23.28 m/s *cos25o = 21.09 m/s The vertical component of velocity vy changes from its initialvalue because there is vetical acceleration.                          vy = vo sin - gt                               = 23.28m/s*sin25 - 9.8m/s2 * 4.35s                               = -32.79 m/s Therefore v = vx2 +vy2                   = 38.98 m/s Angle = -57.25o Angle = -57.25o

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