Question

A ball is thrown into the air with a velocity of 35 ft/s. Its height, in feet, after t seconds is given by
s(t) = 35t − 16t2.
Find the velocity (in ft/s) of the ball at time
t = 1 second.

Answer 1

Answer:

Answer explained below

Step-by-step explanation:

the equation of motion when an object is accelerated is given by:

$x = x_{0}+v_{0}t+\frac{1}{2}at^{2}$ where

x - is the current position(or displacement) of the body

x0 is the initial position

v0 is the initial velocity

a is acceleration

t is time

We rewrite the equation

$s(t) = 35t - 16t^{2}$ which is similar to the equation above, which means we can rewrite the following equation like this:

$s(t) = 35t - \frac{32}{2} t^{2}$

so, in our case:

x0 = 0

v0 = 35 m/s

Now, since this is an accelerated motion it means that the velocity of the body changes as well due to acceleration with the equation:

$v(t) = v_{0}+at$

acceleration is negative, therefore:

$v(t) = 35 - 32t$

t = 1 results:

$v(1) = 35 - 32 = 3 ft/s$