If a cannonball is shot directly upward with a velocity of 160 ft per second, its height above the ground after t seconds is given by s(t)equals160 t minus 16 t squared. Find the velocity and the acceleration after t seconds. What is the maximum height the cannonball reaches? When does it hit the ground? The velocity after t seconds is v(t)equals 160 minus 32 t ft/sec. The acceleration after t seconds is a(t)equals negative 32 ft divided by sec squared. The cannonball reaches a maximum height of 400 ft. The cannonball hits the ground after tequals 10 sec.

Step-by-step explanation:

If a cannonball is shot directly upward with a velocity of 160 ft per second. Its height as a function of time is given by :

.......(1)

t is in seconds

(a) Velocity is given by :

Acceleration is given by :

(b) For maximum height put

i.e.

Put t = 5 s in equation (1). So,

(c) When the ball reaches ground, its height is equal to 0. So,

h(t) = 0

Hence, this is the required solution.

about 0.75 hours

step-by-step explanation:

answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation: