two objects are thrown from the top edge of a cliff with a speed of 10 m/s. one object is thrown straight down and the other straight up. if the first object hits the ground in 4.0 second, the second hits the ground in ________ seconds after the first object.

Answer 1

Assume no air resistance.

For the object thrown downwards,
t=4 s
vi=-10 (+ve upwards)
g=-9.8 m/s²
S=vi*t+(1/2)gt²
=-10*4+(1/2)(-9.8)(4²)
=-118.4 m

For the object thrown downwards,
t=unknown
vi=+10 (+ve upwards)
g=-9.8 m/s²
S=vi*t+(1/2)gt²
-118.4=+10*t+(1/2)(-9.8)(t²) ...(1)

Solving the quadratic equation (1), we get
t=-4 or t=296/49

Can you take it from here?

Answer 2

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Answer 3

12

Answer 4

The object thrown downward takes 4s to hit the ground. The object thrown upward will travel up, then fall, then by the time it reaches its initial position, it will be traveling with a velocity equal in magnitude but opposite in direction to its initial velocity: -10m/s (due to conservation of energy). After that, it will follow the same path that the first ball followed. Therefore, the difference in time from the first ball landing to the second ball landing is equal to the time it took for the second ball to travel up then back down to its initial position. This time can be found using vf = v0 + at. This results in t = 2s. The second ball lands 2s after the first.