Respuesta :
b. -96 ft/sec
For the first part, solve the equation you were given to find the time at which the height of the ball is 0. Sos(t) = -16t^2 - 64t + 800 = -16t^2 - 64t + 80
This is a standard quadratic equation that you can solve using the quadratic formula with a = -16, b = -64, and c = 80, so the roots are: 1 and -5.We can ignore the -5 root since that's what the situation would have looked like if we went back in time 5 seconds and launched the ball upwards from the ground. But since we're not doing time travel, the solution of interest is the 1. So the ball hits the ground one second after throwing it.The formula for distance under constant acceleration is d = 1/2 A T^2. That formula accounts for the -16t^2 term, so A is 32. So during that 1 second it takes for the ball to hit the ground, it will be accelerated 1 * 32 = 32 ft/sec. That is added to the initial velocity the ball was given. So at the moment it hits the ground it's going-64 ft/sec - 32 ft/sec = -96 ft/sec.
For the second problem, the key thing to note is "instantaneous rate of change". When you see that phrase you should immediately think "first derivative". So let's calculate the first derivative of the equation given.S(r)=2πrh+2πr^2S'(r)=2πh+4πr
Now substitute the value 4 for r, givingS'(4)=2πh+4π4 = 2πh+16π
And we have a problem. That value isn't one of the available options, so something is wrong. The problem is choice "b" takes into account the increase in surface area for the end caps on the cylinder, but it doesn't take into account the increase in surface area for the side of the cylinder which is what the 2πh term accounts for. Please show this to your teacher.
For the first part, solve the equation you were given to find the time at which the height of the ball is 0. Sos(t) = -16t^2 - 64t + 800 = -16t^2 - 64t + 80
This is a standard quadratic equation that you can solve using the quadratic formula with a = -16, b = -64, and c = 80, so the roots are: 1 and -5.We can ignore the -5 root since that's what the situation would have looked like if we went back in time 5 seconds and launched the ball upwards from the ground. But since we're not doing time travel, the solution of interest is the 1. So the ball hits the ground one second after throwing it.The formula for distance under constant acceleration is d = 1/2 A T^2. That formula accounts for the -16t^2 term, so A is 32. So during that 1 second it takes for the ball to hit the ground, it will be accelerated 1 * 32 = 32 ft/sec. That is added to the initial velocity the ball was given. So at the moment it hits the ground it's going-64 ft/sec - 32 ft/sec = -96 ft/sec.
For the second problem, the key thing to note is "instantaneous rate of change". When you see that phrase you should immediately think "first derivative". So let's calculate the first derivative of the equation given.S(r)=2πrh+2πr^2S'(r)=2πh+4πr
Now substitute the value 4 for r, givingS'(4)=2πh+4π4 = 2πh+16π
And we have a problem. That value isn't one of the available options, so something is wrong. The problem is choice "b" takes into account the increase in surface area for the end caps on the cylinder, but it doesn't take into account the increase in surface area for the side of the cylinder which is what the 2πh term accounts for. Please show this to your teacher.
