Question

Two identical pellet guns are fired simultaneously from the edge of a cliff. These guns impart an initial speed of 30 m/s to each pellet. Gun A is fired straight upward with the pellet eventually hitting the ground beneath the cliff. Gun b is fired straight downward. Ignore air resistance. Assume g=10 m/s/s.

A. How long after pellet B hits the ground does pellet A hit.

B. If the pellet hits the ground with an impact speed of 60 m/s, how high is the cliff?

Answer 1

a) Since we see that pellet A will come back to the top of the clip with a velocity = -30 m/s , from the cliff down, both the pellets will reach with same time, so the difference is the time it takes for A to go up and come back to the cliff

we have,

V= V₀ + (a*t)

where, V = final speed of pellet ( coming down) = -30 m/s ,

V₀ = initial speed of pellet = 30 m/s,

g = accleration due to gravity = -9.8 m/s²

therefore

-30 = 30 +(-10)*t

implies that time t = 6.00 seconds

b) now if the pellet hits the ground with speed of 60 m/s, then

distance = speed * time = 6.00 * 60 = 360 m