Final answer:
The rock goes to a height of 400 feet. The velocity of the rock when it is 256 feet above the ground on the way up is 32 feet per second. The rock hits the ground after 10 seconds.
Step-by-step explanation:
(a) To calculate the height the rock goes we need to use the equation for height: s = 160t - 16t^2. Since the rock is thrown straight up, it will reach its highest point when its velocity is 0. So, we can find the time it takes to reach the highest point by setting the velocity equation equal to 0 and solving for t: 0 = 160 - 32t. Solving for t gives us t = 5 seconds. Now, we can substitute t = 5 into the height equation to find the height: s = 160(5) - 16(5^2) = 800 - 400 = 400 feet. Therefore, the rock goes to a height of 400 feet.
(b) To find the velocity of the rock when it is 256 feet above the ground on the way up, we can use the velocity equation: v = 160 - 32t. First, we need to find the time it takes the rock to reach a height of 256 feet. We can set the height equation equal to 256 and solve for t: 256 = 160t - 16t^2. Rearranging the equation gives us 16t^2 - 160t + 256 = 0. Solving this quadratic equation gives us two possible values for t: t = 4 and t = 8. However, since we are looking for the time on the way up, we choose t = 4. Now, we can plug t = 4 into the velocity equation to find v: v = 160 - 32(4) = 160 - 128 = 32 feet per second. Therefore, the velocity of the rock when it is 256 feet above the ground on the way up is 32 feet per second.
(c) To find when the rock hits the ground, we can set the height equation equal to 0 and solve for t: 0 = 160t - 16t^2. Rearranging the equation gives us 16t^2 - 160t = 0. Factoring out t gives us t(16t - 160) = 0, so t = 0 or t = 10. Since we are looking for the time when the rock hits the ground, we choose t = 10. Therefore, the rock hits the ground after 10 seconds.