In a Fourth of July celebration, a firecracker is launched from ground level wit
ID: 1707144 • Letter: I
Question
In a Fourth of July celebration, a firecracker is launched from ground level with an initial velocity of 28.6 at 39.1 from the vertical. At its maximum height it explodes in a starburst into many fragments, two of which travel forward initially at 30.0 at 59.9 with respect to the horizontal, both quantities measured relative to the original firecracker just before it exploded.With what angles with respect to the horizontal do the two fragments initially move right after the explosion, as measured by a spectator standing on the ground?Explanation / Answer
The original firecracker moves as a projectile .At its maximum height its velocity is horizontal . The velocity of fragment A relative to the ground is related to the velocity of original fire cracker relative to the ground and the velocity of fragment relative to fire cracker v A/G = v A/ F + v F/G Fragment B also obeys a similar equation Let x be the direction of horizontal motion of the fire cracker before it explodes and y be up wards . Fragment A moves at 59.9 degrees above x direction and fragment B moves 59.9 degrees below x direction Before it explodes the fire cracker has a x = 0 , ay = -9.8 m / s2 The horizontal component of firecrackers velocity relative to ground v F/Gx = 28.6 cos 39.1 = 22.194 m/s at the time of explosion vF/Gy = 0 For fragment A vA/Fx = 30 cos 59.9 = 15.045 m/s v A/Fy = 30 sin 59.9 = 25.954 m/s v A/Gx = vA/Fx + v F/Gx = 15.045 + 22.194 = 37.234 m/s v A/Gy = v A/Fy = 25.954 m/s tan o = 25.954/37.234 = 34.878 degrees the frragments move at 34.878o above and 34.878o below the horizontalRelated Questions
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