Calculus Help- Vertical Motion?

Question

The height of of an object moving vertically is given by

s = -16t^2 + 96t + 112

with s in ft and t in sec. Find (a.) the object's velocity when t=0, (b.) its maximum height and when it occurs, and (c.) its velocity when s=0.

Answer 1

s = ... could be called a position function.
The velocity is the change in postition with respect to time. For any particular moment, we would be trying to find the instantaneous velocity. It is for this that calculus is required.
We can differentiate both sides:
ds/dt = -32t + 96.
So at t=0, ds/dt is -32*0 + 96 = 96.
So the velocity is 96 ft/s.

Observe that the position function, when plotted as a graph, forms an upside-down parabola. The peak of this figure is the maximum height of your object. Observe the slope (rate of change) of the peak is zero, i.e., the velocity is zero. So let ds/dt=0:
0 = -32t + 96,
t = 96/32 = 3 seconds.

When s=0, we have:
0 = -16t^2 + 96t + 112.
Solve for t and plug that value into:
ds/dt = -32t + 96
to get the velocity.

Feel free to email me for clarifications.

Answer 2

confident. Your attitude is right. velocity = ds/dt = -2t + ninety six s = 0 => t^2 - 96t - 112 = 0 => t = (a million/2) [ ninety six ± ?[(ninety six)^2 + 448]] => t = (a million/2) [ ninety six ± ninety 8.31] => t = - a million.sixteen s or 97.sixteen s Any value of t could desire to provide the comparable value with opposite indicators. t = - a million.sixteen => v = -2(-a million.sixteen) + ninety six = ninety 8.32 feet/s t = 97.sixteen => v = -2 (97.sixteen) + ninety six = -ninety 8.32 feet/s At a million.sixteen sec in the previous than while the item substitute into at a top of 112 feet, it substitute into thrown upwards from s = 0 with a velocity of ninety 8.32 feet/s and alter into lower back to s = 0 shifting downwrds with a velocity of ninety 8.32 feet/s.