Answer :
Final answer:
The rock's velocity as it hits the bottom of the hole, after being tossed up with a velocity of 35.8 m/s and falling into a 12.5 m deep hole, is approximately 39.1 m/s. This result is found by using the kinematic equation considering gravitational acceleration and the additional distance fallen.
Explanation:
When the rock is tossed straight up with a velocity of 35.8 m/s, it will rise until its velocity decreases to zero due to gravity, then it will fall back down. The rock's velocity upon returning to its starting height will be the same as it was thrown up but in the opposite direction, as per kinematic equations for gravitational acceleration. However, since the rock continues to fall into the hole and goes 12.5 m deeper, we need to calculate the additional velocity gained during this fall.
To calculate this, we can use the kinematic equation:
v^2 = u^2 + 2as
Where v is the final velocity, u is the initial velocity (which is 35.8 m/s downwards when it just starts to fall back), a is the acceleration due to gravity (9.8 m/s^2), and s is the distance fallen (12.5 m).
Therefore, the velocity of the rock as it hits the bottom of the hole is:
v = √(u^2 + 2as)
= √((35.8 m/s)^2 + 2 * 9.8 m/s^2 * 12.5 m)
Plugging in the values, we get:
v = √(1281.64 + 245)
= √(1526.64)
≈ 39.1 m/s
Therefore, the rock's velocity as it hits the bottom of the hole is approximately 39.1 m/s.
