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?

Respuesta :

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:

[tex]v(t) = \frac{d}{dt}h(t)[/tex] (1)

Where [tex]v(t)[/tex] is the velocity of the object at time [tex]t[/tex], in feet per seconds.

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

[tex]v(t) = -32\cdot t+64[/tex]

[tex]-32\cdot t + 64 = 0[/tex]

[tex]t = 0.5\,s[/tex]

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