Two identical pellet guns are fired simultaneously from the edge of a cliff. These guns impart an initial speed of 44.7 m/s to each pellet. Gun A is fired straight upward, with the pellet going up and then falling back down, eventually hitting the ground beneath the cliff. Gun B is fired straight downward. In the absence of air resistance, how long after pellet B hits the ground does pellet A hit the ground?
2 answers
Bullet A hits the ground after 9 sec.
Gun A:
V = Vo + g*Tr = 0.
Tr = -Vo/g = -44.7/-9.8 = 4.56 s. = Rise
time.
h = Vo*Tr + 0.5g*Tr^2 = 44.7*4.56 - 4.9*4.56^2 = 102 m.
Tf = Tr = 4.56 s. = Time to fall back
to top of cliff.
T = Tr+Tf = 4.56 + 4.56 = 9.12 s. = Time
to rise and fall back to top of cliff.
The time required for each pellet to
fall from top of cliff to ground is
equal. Therefore, pellet A hits ground
9.12 s after pellet B.
V = Vo + g*Tr = 0.
Tr = -Vo/g = -44.7/-9.8 = 4.56 s. = Rise
time.
h = Vo*Tr + 0.5g*Tr^2 = 44.7*4.56 - 4.9*4.56^2 = 102 m.
Tf = Tr = 4.56 s. = Time to fall back
to top of cliff.
T = Tr+Tf = 4.56 + 4.56 = 9.12 s. = Time
to rise and fall back to top of cliff.
The time required for each pellet to
fall from top of cliff to ground is
equal. Therefore, pellet A hits ground
9.12 s after pellet B.
4 answers