Final answer:
The height of the cliff is 8.55 m, which is calculated by using the kinematic equation with an initial upward velocity of 8 m/s and a time of 2.35 s. The time it takes the rock to hit the ground when thrown down with the same speed is also 2.35 seconds since gravity and initial velocity work together to shorten the time.
Step-by-step explanation:
To calculate the height of a cliff based on the time it takes for a rock to hit the ground when thrown up with an initial velocity, we use the kinematic equation for projectile motion:
s = ut + \frac{1}{2}at^2 where s is the displacement, u is the initial velocity, a is the acceleration due to gravity (9.81 m/s^2), and t is the time. Since the rock was thrown up, the initial velocity is positive.
The rock will go up and then come down, so the total distance covered will be the height of the cliff plus the additional height it climbed before falling back down. Let's substitute the given values into the equation: s = (8.00 m/s)(2.35 s) + \frac{1}{2}(-9.81 m/s^2)(2.35 s)^2. This gives us s = 8.55 m, which is choice (a).
When the rock is thrown straight down with the same speed, gravity will assist the rock's downward motion, increasing the speed at which it hits the ground. Therefore, the time to reach the ground will be less than when it was thrown upwards. Since both gravity and initial velocity work in the same direction in this case, the equation can be simplified to: h = ut + \frac{1}{2}at^2 where h is the cliff height, u is the initial downward velocity, and t is the time. Solving this equation for time gives us t = 2.35 s, which is choice (a).