s(t) = 16 + 24t − 16t2

Question

How long does it take (in seconds) for the ball to get back to the platform?

v(t)= -32t +24

i know you set velocity equal to zero but it’s not accepting my answer of 0.75.

Answer 1

When your launch the ball upward at whatever initial velocity, at some point (due to gravity and air resistance) it will be slowed to a velocity of 0. This happens at the height of the arch. Then gravity will accelerate it downward. When you set velocity to zero, this finds the time of half the journey (from ground to top). Double this to find the total time. Assuming you have copied the problem down correctly, double .75 to get 1.5 seconds.

Answer 2

Assuming the ball is thrown straight up. Actually, the velocity is zero at the maximum height of a parabola, when the ball starts falling back down toward the platform. What you want is when the distance is zero, or s(t) = 0.

-16t²+24t+16 = 0

Dividing through by -16:

t²-1.5t-1 = 0

Using the quadratic equation:

t = (1.5±√(2.25+4))/2 = 2, -1/2

Choose the positive value or t = 2 seconds.

This is the time for the ball to reach the ground. The problem is asking for the time to get back to the platform of 16.

-16t²+24t+16 = 16

t = 1.5 seconds

which is double the time it takes for the ball to reach the platform height.

Answer 3

Setting v = 0 gets you to the peak of the trajectory. for the full trajectory, what you want is v = -v₀ = -24

If you do that you will find that that will be double the time that you calculated.

Please consider a tutor. Take care.

Answer 4