Question

When an object is thrown upwards with a speed of 64 ft/sec, its height above the ground is given by the function h(t)=−16t2+64t , where t is the time in seconds after it has been thrown. At what time will the object reach its highest point?

Answer 1

Answer:

The object will reach its highest point 0.5 seconds after it has been thrown.

Step-by-step explanation:

The object reaches its maximum height when velocity is equal to zero, the velocity is the derivative of function height. That is:

$v(t) = \frac{d}{dt}h(t)$ (1)

Where $v(t)$ is the velocity of the object at time $t$, in feet per seconds.

If we know that $h(t) =-16\cdot t^{2}+64\cdot t$ and $v(t) = 0\,\frac{ft}{s}$, then the time when object reaches its highest point is:

$v(t) = -32\cdot t+64$

$-32\cdot t + 64 = 0$

$t = 0.5\,s$

The object will reach its highest point 0.5 seconds after it has been thrown.