a rock is thrown witha velocity of 20 meters per second from the top of a 24 meter high cliff and it misses the cliff on the way back down. when will the rock be 10 meters from the water below? round your answer to two decimal places
To find out when the rock will be 10 meters from the water below, we can use the equations of motion.
Let's break down the problem step by step:
1. Determine the initial vertical velocity (v) of the rock when it is thrown from the top of the cliff. In this case, the rock is thrown upwards, so we'll take the velocity as positive. Therefore, v = +20 m/s.
2. Determine the initial vertical displacement (y₀) of the rock when it is thrown from the top of the cliff. Here, the rock is thrown from a height of 24 meters, so y₀ = +24 m.
3. Determine the gravitational acceleration (g). On Earth, the acceleration due to gravity is approximately 9.8 m/s².
4. Calculate the time it takes for the rock to reach the maximum height (t_peak). We can use the equation:
v = u + at
where v is the final velocity (which is 0 when the rock reaches the maximum height), u is the initial velocity, a is acceleration, and t is time.
Plugging in the known values, we have:
0 = 20 - 9.8t_peak
Solving for t_peak gives:
t_peak = 2.04 s (rounded to two decimal places)
5. Calculate the vertical displacement (y_peak) at the maximum height using the equation:
y_peak = y₀ + u × t_peak - 1/2 × g × t_peak²
Plugging in the known values, we have:
0 = 24 + 20 × 2.04 - 1/2 × 9.8 × (2.04)²
Solving for y_peak gives:
y_peak = 44.04 m (rounded to two decimal places)
6. Calculate the time it takes for the rock to reach a vertical displacement of 10 meters below the maximum height (t_final). We can use the equation:
y = y_peak + u × t_final - 1/2 × g × t_final²
Plugging in the known values, we have:
-10 = 44.04 + 20 × t_final - 1/2 × 9.8 × t_final²
This equation is quadratic in form, so we can solve it using the quadratic formula:
t_final = (-b ± √(b²-4ac))/(2a)
where a = -4.9 (1/2 × g), b = 20, and c = 54.04 (44.04 - 10).
Plugging in the values and solving for t_final gives:
t_final = 2.42 s (rounded to two decimal places) or t_final = -5.62 s
Since time cannot be negative in this context, we disregard the negative value.
Therefore, the rock will be 10 meters from the water below approximately 2.42 seconds after it was thrown from the top of the cliff.