Before you start working on this one, it's important to understand a few things about what's actually going on.
-- Horizontal speed and vertical speed have no effect on each other. In Physics problems, you handle them separately.
-- Throwing an object horizontally, like in this problem, has no effect on how long it takes to reach the ground.
-- In this problem, the 10 m/s is horizontal. It has no effect on the vertical motion of the ball, or on how long the ball takes to reach the ground.
-- Rule to understand and memorize:
. . . . . Only gravity affects vertical speed. Horizontal speed doesn't.
. . . . . Gravity affects only vertical speed, not horizontal speed.
-- The ball's VERTICAL speed starts from zero, as if it were simply dropped, or just rolled off the edge of the cliff.
-- So when you start working on this problem, you ignore the 10 m/s for a while, and just calculate how long it takes a ball to ht the ground if it's DROPPED from 45 meters.
The distance a dropped object falls in time 't' is D= 1/2 g t² . We know 'D', and we know 'g'.
45m = 1/2 (9.8 m/s²) t²
45m = (4.9 m/s²) t²
Divide both sides by 4.9 m/s² :
t² = (45m) / (4.9 m/s²)
t² = 9.18 sec²
Square root both sides"
t = 3.03 seconds
That's the time it takes the ball to reach the ground, no matter what horizontal speed it had when it started to fall ... just as long as it had no initial vertical speed.
So why does this problem even tell us about the 10 m/s horizontal speed ?
We need that for the second part of the question: How far away from the base of the cliff does it hit the ground ?
The ball was thrown horizontally at 10 m/s. Gravity has no effect on horizontal speed, so the ball just keeps moving horizontally at 10 m/s. But it has to stop doing that after 3.03 seconds, because at that time, it is surprised to find that it has hit the ground.
In that 3.03 seconds, the ball has traveled (10 m/s)x(3.03 s) = 30.3 m horizontally, from the base of the cliff.
