Answer :
Final answer:
To calculate the rock's velocity as it hits the bottom of the hole, kinematic equations are used to first determine the time taken for the rock to reach the highest point and return to the original height. The additional time to fall into the hole is then added, and the final velocity of the rock at the bottom of the hole is calculated using the acquired total time.
Explanation:
To calculate the rock's velocity as it hits the bottom of the hole, we can use kinematic equations for uniformly accelerated motion. Since the problem states to ignore air resistance, we'll consider the only acceleration acting on the rock to be due to gravity (9.8 m/s2). When the rock is tossed straight up at 35.8 m/s, it will slow down under gravity's pull, come to a stop, and then fall back down.
Firstly, we need to find the time it takes for the rock to come back down to the original height from which it was thrown. This can be done using the equation:
v = u + at
Where:
- v is the final velocity (0 m/s at the highest point)
- u is the initial velocity (35.8 m/s upwards)
- a is the acceleration due to gravity (-9.8 m/s2, negative because it's in the opposite direction of the initial throw)
- t is the time in seconds
The time taken to reach the highest point t, can be calculated. We then double this time to account for the rock’s ascent and descent back to the original height. Next, we calculate how long the rock would take to fall an additional 12.5 m into the hole using the equation:
s = ut + 1/2at2
We add this time to the time taken for the rock to return to the original height and then use the first kinematic equation to find the velocity (v) at the bottom of the hole. Remember to adjust the sign of the velocity, as it will be directed downwards.
